Tìm x, biết:
d.\(\left(\frac{3}{4}\right)^x=\frac{2^8}{3^4}\)
c., \(3^x.3^{12}=9^2.81^5\)
e. \(5.8^x-2^{3x+1}=192\)
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)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
Giải các phương trình sau:
a,\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
b,\(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
c,\(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
d,\(\frac{x-4}{5}+\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
e,\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
f,\(\frac{3x-1}{2}-\left(x-\frac{1}{4}\right)=\frac{4x-9}{8}\)
a, Ta có : \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)
=> \(3\left(2x-1\right)-5\left(x-2\right)=x+7\)
=> \(6x-3-5x+10-x-7=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
b, Ta có : \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
=> \(\frac{3\left(x+3\right)}{6}-\frac{2\left(x-1\right)}{6}=\frac{x+5}{6}+\frac{6}{6}\)
=> \(3\left(x+3\right)-2\left(x-1\right)=x+5+6\)
=> \(3x+9-2x+2-x-5-6=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(\frac{2\left(x+5\right)}{3}+\frac{x+12}{2}-\frac{5\left(x-2\right)}{6}=\frac{x}{3}+11\)
=> \(\frac{4\left(x+5\right)}{6}+\frac{3\left(x+12\right)}{6}-\frac{5\left(x-2\right)}{6}=\frac{2x}{6}+\frac{66}{6}\)
=> \(4\left(x+5\right)+3\left(x+12\right)-5\left(x-2\right)=2x+66\)
=> \(4x+20+3x+36-5x+10-2x-66=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
a, (x-1)3 - x(x-1)2 = 5(2-x) - 11(x+2)
b, (x-2)3 + (3x-1)(3x+1) = (x+1)3
c, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
d, \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
e, \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)
⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)
⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)
⇔\(9x-10=0\)
hay 9x=10
⇔\(x=\frac{10}{9}\)
Vậy: \(x=\frac{10}{9}\)
c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)
⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)
⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)
⇔\(6x-3-5x+10-3x-21=0\)
⇔\(-2x-14=0\)
⇔\(-2x=14\)
hay x=-7
Vậy: x=-7
d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
⇔\(6x-18+7x-35-13x-4=0\)
⇔\(-21\ne0\)
Vậy: x∈∅
e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)
⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)
⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)
⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)
⇔\(-12x+96=0\)
⇔\(-12x=-96\)
hay x=8
Vậy: x=8
tìm số x , biết :
a) \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
b) \(\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=\frac{-53}{10}\)
c) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
d) \(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
e) \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
f) \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{2}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{2}\)
\(-\frac{5}{6}\times x=\frac{5}{2}\)
\(x=\frac{5}{2}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{2}\times\left(-\frac{6}{5}\right)\)
\(x=-3\)
b.
\(\frac{2}{5}+\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{53}{10}-\frac{2}{5}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=\frac{-53-4}{10}\)
\(\frac{3}{5}\times\left(3x-3,7\right)=-\frac{57}{10}\)
\(3x-3,7=-\frac{57}{10}\div\frac{3}{5}\)
\(3x-3,7=-\frac{57}{10}\times\frac{5}{3}\)
\(3x-\frac{37}{10}=-\frac{19}{2}\)
\(3x=-\frac{19}{2}+\frac{37}{10}\)
\(3x=\frac{-95+37}{10}\)
\(3x=-\frac{58}{10}\)
\(3x=-\frac{29}{5}\)
\(x=-\frac{29}{5}\div3\)
\(x=-\frac{29}{5}\times\frac{1}{3}\)
\(x=-\frac{29}{15}\)
c.
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23-15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\times\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21-16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}\times\frac{4}{3}\)
\(x=\frac{5}{6}\)
d.
\(-\frac{2}{3}\times x+\frac{1}{5}=\frac{3}{10}\)
\(-\frac{2}{3}\times x=\frac{3}{10}-\frac{1}{5}\)
\(-\frac{2}{3}\times x=\frac{3-2}{10}\)
\(-\frac{2}{3}\times x=\frac{1}{10}\)
\(x=\frac{1}{10}\div\left(-\frac{2}{3}\right)\)
\(x=\frac{1}{10}\times\left(-\frac{3}{2}\right)\)
\(x=-\frac{3}{20}\)
e.
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{20+9}{12}\)
\(\left|x\right|=\frac{29}{12}\)
\(x=\pm\frac{29}{12}\)
Vậy \(x=\frac{29}{12}\) hoặc \(x=-\frac{29}{12}\)
f.
\(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)
\(\left|2x-\frac{1}{3}\right|=1-\frac{5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{6-5}{6}\)
\(\left|2x-\frac{1}{3}\right|=\frac{1}{6}\)
\(2x-\frac{1}{3}=\pm\frac{1}{6}\)
\(2x-\frac{1}{3}=\frac{1}{6}\)\(2x=\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{1+2}{6}\)
\(2x=\frac{3}{6}\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{2}\div2\)
\(x=\frac{1}{2}\times\frac{1}{2}\)
\(x=\frac{1}{4}\)
\(2x-\frac{1}{3}=-\frac{1}{6}\)\(2x=-\frac{1}{6}+\frac{1}{3}\)
\(2x=\frac{-1+2}{6}\)
\(2x=\frac{1}{6}\)
\(x=\frac{1}{6}\div2\)
\(x=\frac{1}{6}\times\frac{1}{2}\)
\(x=\frac{1}{12}\)
Vậy x = 1/4 hoặc x = 1/12.
Chúc bạn học tốt
Sorry nha, mik chép lộn đềLàm lại câu a nha
a.
\(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\left(\frac{2}{3}-\frac{3}{2}\right)\times x=\frac{5}{12}\)
\(\left(\frac{4-9}{6}\right)\times x=\frac{5}{12}\)
\(-\frac{5}{6}\times x=\frac{5}{12}\)
\(x=\frac{5}{12}\div\left(-\frac{5}{6}\right)\)
\(x=\frac{5}{12}\times\left(-\frac{6}{5}\right)\)
\(x=-\frac{1}{2}\)
Chúc bạn học tốt
\(a,\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\)
\(\frac{4}{6}x-\frac{9}{6}x=\frac{5}{12}\)
\(\frac{-5}{6}x=\frac{5}{12}\)
\(x=\frac{5}{12}:\frac{-5}{6}\)
\(x=\frac{-1}{2}\)
\(b,\frac{2}{5}+\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}\)
\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{2}{5}\)
\(\frac{3}{5}\cdot\left(3x-3,7\right)=\frac{-53}{10}-\frac{4}{10}\)
\(3x-3,7=\frac{-57}{10}:\frac{3}{5}\)
\(3x=\frac{-19}{2}+\frac{37}{10}\)
\(3x=\frac{-95}{10\text{}\text{}\text{}}+\frac{37}{10}\)
\(x=\frac{-29}{5}:3\)
\(x=\frac{-29}{15}\)
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 !
Tìm x:
a)\(\left(2x+3\right)^2=\frac{9}{21}\)
b)\(\left(3x-1\right)^3=\frac{-8}{27}\)
c)\(x^{10}=25.x^8\)
d)\(\left(x^4\right)^2=\frac{x^{12}}{x^{15}}\)
e)\(\left(x+3\right)^3=\left(x+3\right)^5\)
Cầu giúp đỡ~
e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
Tìm \(x,\) biết:
a) \(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)
b) \(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x+9\right|=5\)
c) \( \left|2\frac{1}{5}-x\right|+\left|x-\frac{1}{5}\right|+8\frac{1}{5}=1,2\)
d) \(2\left|x+3\frac{1}{2}\right|+\left|x\right|-3\frac{1}{2}=\left|2\frac{1}{5}-x\right|\)
Bài 4.1: Tìm x, biết
a) \(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)
b) \(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
c) \(\left|2\frac{1}{5}-x\right|+\left|x-\frac{1}{5}\right|+8\frac{1}{5}=1,2\)
d) \(2\left|x+3\frac{1}{2}\right|+\left|x\right|-3\frac{1}{2}=\left|2\frac{1}{5}-x\right|\)