A) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
B) \(\left(2,4x-36\right):1\frac{5}{7}=-1\)
C) \(\left(3\frac{4}{5}-2x\right).1\frac{1}{3}=5\frac{5}{7}\)
Cần gấp ạ !! Mong mọi người giúp đỡ !!
A) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
B) \(\left(2,4x-36\right):1\frac{5}{7}=-1\)
C) \(\left(3\frac{4}{5}-2x\right).1\frac{1}{3}=5\frac{5}{7}\)
Cần gấp ạ!!!Mong moi người giúp ạ !!
a) \(\frac{3}{x-5}=\frac{-4}{x+2}\)(ĐKXĐ: \(x\ne5;x\ne-2\))
\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)(thỏa mãn ĐKXĐ)
b) \(2,4x-36=-\frac{7}{12}\)
\(\Leftrightarrow2,4x=\frac{425}{12}\)
\(\Leftrightarrow x=\frac{2125}{144}\)
c) \(\left(\frac{19}{5}-2x\right).\frac{4}{3}=\frac{40}{7}\)
\(\Leftrightarrow\frac{19}{5}-2x=\frac{30}{7}\)
\(\Leftrightarrow2x=-\frac{17}{35}\)
\(\Leftrightarrow x=-\frac{17}{70}\)
a)\(\frac{-3}{2}-2x+\frac{3}{4}=-2\)
b)\(\left(\frac{-2}{3}x-\frac{3}{5}\right)\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
c)\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)
d)\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
e)\(\frac{2}{3x}-\frac{3}{12}=\frac{4}{5}-\left(\frac{7}{x}2\right)\)
k)\(\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
m)\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
n)\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Giải giúp mk
mk đang cần ghấp trong hôm nay
cám ơn mọi người
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
\(A=2\frac{3}{7}:\left(-12\frac{1}{2}+15\frac{1}{7}\right).\left(3\frac{3}{10}\right)-\left(\frac{-34}{37}\right).\left(2\frac{1}{10}-4\frac{2}{5}\right)\)
\(B=-3\frac{1}{10}:\left(\frac{1}{2}-\frac{2}{5}\right)-2\frac{1}{28}:\left(\frac{2}{7}-\frac{3}{4}\right)\)
bài 2:Tìm x
a. \(-1\frac{2}{3}-\left|2x-1\right|:\frac{3}{5}=-2\)
b. \(\frac{x-7}{36}=\frac{-4}{7-x}\)
Mình đang cần gấp ai đúng mình tích cho
Mia mình nộp rồi
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
Tìm x biết:
a) \(\left(x+\frac{1}{2}\right).\left(x-\frac{3}{4}\right)=0\)
b) \(\left(\frac{1}{2}.x-3\right).\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
c) \(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
d) \(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
MONG CÁC BN GIÚP ĐỠ MK BÀI NÀY , MK ĐANG CẦN RẤT GẤP GIẢI CHI TIẾT RA GIÚP MK VS NHÉ !!!MK RẤT CẢM ƠN!
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
1) Giải các pt sau:
a) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
b) \(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
c) \(\frac{x+8}{6}-\frac{2x-5}{5}=\frac{x-1}{3}-x+7\)
d) \(\frac{7x}{8}-5\left(x-9\right)=\frac{2x+1,5}{6}\)
e) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
f) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
a) \(\frac{7x-3}{x-1}=\frac{2}{3}\)
b) \(\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
c) \(\frac{8-x}{x-7}-8=\frac{1}{x-7}\)
d) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
e) \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
f) \(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
Mong mọi người giúp đỡ >~~< >__<
a)
\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)
Vậy \(x=\left\{\frac{7}{19}\right\}\) là nghiệm của phương trình.
b)
\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3x-6}{x-2}-\frac{3-x}{x-2}=0\\ \Leftrightarrow1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Rightarrow x=\frac{8}{4}=2\)
Mà \(ĐKXĐ:x\ne2\\ \Rightarrow x\in\varnothing\)
Hay phương trình vô nghiệm.
c)
\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8x-56}{x-7}-\frac{1}{x-7}=0\\ \Leftrightarrow8-x-8x+56-1=0\\ \Leftrightarrow63-9x=0\\ \Rightarrow x=\frac{63}{9}=7\)
Mà \(ĐKXĐ:x\ne7\\ \Rightarrow x\in\varnothing\)
Hay phương trình vô nghiệm.
d)
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\\ \Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2-20=0\\ \Leftrightarrow x^2+10x+25-x^2+10x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Rightarrow x=1\)
Vậy \(x=1\) là nghiệm của phương trình.
e)
\(\frac{x}{2\cdot\left(x-3\right)}+\frac{x}{2\cdot\left(x+1\right)}=\frac{2x}{\left(x+1\right)\cdot\left(x-3\right)}\\ \Leftrightarrow\frac{x^2+x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}+\frac{x^2-3x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}-\frac{4x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Mà \(ĐKXĐ:x\ne\left\{-1;3\right\}\\ \Rightarrow x=0\)
Vật \(x=0\) là nghiệm của phương trình.
f)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\\ \Leftrightarrow\frac{12x+120}{12x\cdot\left(x+10\right)}+\frac{12x}{12x\cdot\left(x+10\right)}-\frac{x^2+10x}{12x\cdot\left(x+10\right)}=0\\ \Leftrightarrow12x+120+12x-x^2-10x=0\\ \Leftrightarrow14x+120-x^2=0\\ \Leftrightarrow x^2-14x-120=0\\ \Leftrightarrow x^2+6x-20x-120=0\\ \Leftrightarrow x\cdot\left(x+6\right)-20\cdot\left(x+6\right)=0\\ \Leftrightarrow\left(x-20\right)\cdot\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-20=0\\x+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)
Vậy \(x=\left\{20;-6\right\}\) là ngiệm của phương trình.
a) \(\frac{1}{7}+\frac{6}{7}:\frac{3}{7}\)
b) \(\frac{4}{5}-\frac{1}{5}.\left(-3\right)\)
c)\(\frac{3}{7}+\left(\frac{-5}{2}\right)-\left(-\frac{3}{5}\right)\)
d) \(\left(\sqrt{25}+\sqrt{49}\right).\frac{1}{4}-|-3|5)12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
Mong mng giúp e vs ạ!!!
a) \(\frac{1}{7}+\frac{6}{7}:\frac{3}{7}\)
\(=\frac{1}{7}+\frac{6}{7}.\frac{7}{3}\) (nhân nghịch đảo)
\(=\frac{1}{7}+2\)
\(=\frac{15}{7}\)
b) \(\frac{4}{5}-\frac{1}{5}.\left(-3\right)\)
\(=\frac{4}{5}-\left(-\frac{3}{5}\right)\)
\(=\frac{7}{5}\)
c) \(\frac{3}{7}+\left(\frac{-5}{2}\right)-\left(-\frac{3}{5}\right)\)
\(=\frac{3}{7}-\left(-\frac{5}{2}\right)+\frac{3}{5}\)
\(=\frac{30}{70}+\frac{175}{70}+\frac{42}{70}\)
\(=\frac{30+175+42}{70}\)
\(=\frac{247}{70}\)
d) viết lại đề hộ mình nhé
Giúp mk vs đg cần gấp nha cảm ơn mấy bạn nhiều lắm !!!
\(\left(\frac{1}{2}\right)^5.x=\left(\frac{1}{2}\right)^7\)
\(\left(\frac{3}{7}\right)^2.x=\left(\frac{9}{21}\right)^2\)
Bài tìm x biết :
2x=2
3x=34
7x=77
(-3)x=(-3)5
(-5)x=(-5)4
2x=4
2x=8
2x=16
3x+1=32
5x-1=5
6x+4=610
52x-7=511
(-2)4x+2=64
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\)
\(\left(-\frac{1}{10}\right)^x=\frac{1}{100}\)
\(\left(-\frac{3}{2}\right)^x=\frac{9}{4}\)
\(\left(-\frac{3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(-\frac{2}{3}\right)^x=-\frac{8}{27}\)
nhiều quá trời, lm chắc mỏi tay luôn
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe. đánh tới què tay, hoa mắt lun r nekkk!!
người kiểu gì mà bài này ko làm đc