Tìm x biết :
a) \(\frac{x-2}{16}=\frac{-4}{2-x}\)
b) \(\left(2x+7\right)^2-28=64\)
Tìm x biết
a.\(|x|=\frac{4}{7},x>0\)
b.(2x=3)\(^2\)=64
c.\(\left(\frac{1}{2}\right)^x=\frac{1}{16}\)
d.\(3^{x+1}=27\)
\(\left|x\right|=\frac{4}{7}\)
\(\Rightarrow x=\frac{4}{7}\)
b,\(\left(2x-3\right)^2=64\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm8\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=8\\2x-3=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{5}{2}\end{cases}}}\)
c,\(\left(\frac{1}{2}\right)^x=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{2}\right)^x=\left(\pm\frac{1}{2}\right)^4\)
\(\Rightarrow x=\pm\frac{1}{2}\)
d,\(3^{x+1}=27\)
\(\Leftrightarrow3^{x+1}=3^3\)
\(\Leftrightarrow x+1=3\)
\(\Leftrightarrow x=2\)
Tìm x biết: \(\frac{4}{\left(x+2\right).\left(x+6\right)}+\frac{7}{\left(x+6\right).\left(x+13\right)}=\frac{2x+1}{\left(x+2\right).\left(x+16\right)}-\frac{3}{\left(x+13\right).\left(x+16\right)}\)
Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)
<=>1/x+2 -1/x+6 +1/x+6 -1/x+13
<=>1/x+2-1/x+13
=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)
<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)
<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)
<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)
<=> 2x+1=14
<=> 2x=14-1
<=> 2x=13
<=> x=13:2
<=> x=13/2
Vậy x=13/2
Chúc bạn học tốt
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
1a)tìm x,y biết: \(4+\frac{x}{7+y}=\frac{4}{7}and:x+y=22\)
b)cho \(\frac{x}{3}=\frac{y}{4}\)và \(\frac{y}{5}=\frac{z}{6}\). Tính M=\(\frac{2x+3y+4z}{3x+4y+5z}\)
c) tìm x biết \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=2^x\)
d)\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
2. Tính:P=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+..+16\right)\)
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
2.
Ta có 1+2+...+n=n.(n+1):2
=>P=\(1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\)\(\frac{1}{16}.\frac{16.17}{2}\)=1+\(\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)=1+\(\frac{1}{2}.\left(3=4+..=17\right)\)
=1+\(\frac{1}{2}.153=1+\frac{153}{2}=\frac{155}{2}\)
Tìm x biết rằng
a)\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}...\frac{30}{62}.\frac{31}{64}=2x\)
b)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
Tìm x biết : \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}...\frac{30}{62}.\frac{31}{64}=2x\)
Biến đổi mẫu thức của vế trái : \(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}...\frac{30}{2.31}.\frac{31}{2.32}=2x\) Giản ước vế trái đực :
\(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2.32}=2x\) (*)
\(\Leftrightarrow\frac{1}{2^{36}}=2x\Leftrightarrow x=\frac{1}{2^{37}}\)
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
4. Tìm x biết:
a) (3x - 7)5 = 0;
b) \(\frac{1}{4}\) - (2x - 1)2 = 0;
c) \(\frac{1}{16}\) - (5 - x)3 = \(\frac{31}{64}\);
d) 2x = (3,8)3 : (-3,8)2;
e) \(\left(\frac{27}{64}\right)^9\). x = \(\left(-\frac{3}{4}\right)^{32}\);
f) \(5^{\left(x+5\right)\left(x^2-4\right)}\) = 1;
g) (x - 2,5)2 = \(\frac{4}{9}\);
h) \(\left(2x+\frac{1}{3}\right)^3=\frac{8}{27}\)
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
Tìm x biết: \(\frac{4}{\left(x+2\right).\left(x+6\right)}+\frac{7}{\left(x+6\right).\left(x+13\right)}=\frac{2x+1}{\left(x+2\right).\left(x+16\right)}-\frac{3}{\left(x+13\right).\left(x+16\right)}\)
Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)
<=>1/x+2 -1/x+6 +1/x+6 -1/x+13
<=>1/x+2-1/x+13
=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)
<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)
<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)
<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)
<=> 2x+1=14
<=> 2x=14-1
<=> 2x=13
<=> x=13:2
<=> x=13/2
Vậy x=13/2
Chắc là vầy. Mk cug ko chắc nữa
Bài 1: Thực hiện các phép tính dau bằng cách hợp lí
a. \(\frac{11}{225}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
b. \(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
Bài 2: Tìm x biết
a. \(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
b. \(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
Bài 3: Thực hiện các phép tính sau bằng cách hợp lí nhất
a. \(\left(-\frac{40}{51}\cdot0,32\cdot\frac{17}{20}\right):\frac{64}{75}\)
b. \(-\frac{10}{11}\cdot\frac{8}{9}+\frac{7}{18}\cdot\frac{10}{11}\)
c. \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{29}{42}-8\)
d. \(-1\frac{5}{7}\cdot15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
Bìa 4: Tính giá trị của các biểu thức sau
a. \(A=7x-2x-\frac{2}{3}y+\frac{7}{9}y\) với \(x=-\frac{1}{10};y=4,8\)
b. \(B=x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}}\) với\(x=-\frac{1}{3}\)