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a) \(\left(4x-13\right)^4+4^3=145\)
\(\Rightarrow\left(4x-13\right)^4+64=145\)
\(\Rightarrow\left(4x-13\right)^4=81\)
\(\Rightarrow4x-13=\pm3\)
+) \(4x-13=3\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
+) \(4x-13=-3\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=4\) hoặc \(x=\frac{5}{2}\)
b) \(3^{x+2}-3^x=72\)
\(\Rightarrow3^x.3^2-3^x=72\)
\(\Rightarrow3^x.\left(3^2-1\right)=72\)
\(\Rightarrow3^x.8=72\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
c) \(2^{x+2}-2^{x-1}=224\)
\(\Rightarrow2^{x-1+3}-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.2^3-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.\left(2^3-1\right)=224\)
\(\Rightarrow2^{x-1}.7=224\)
\(\Rightarrow2^{x-1}=32\)
\(\Rightarrow2^{x-1}=2^5\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
Vậy x = 6
a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)
hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)
Vậy: \(x=\frac{9}{10}\)
b) Ta có: \(5\frac{4}{7}:x=13\)
\(\Leftrightarrow\frac{39}{7}:x=13\)
\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)
Vậy: \(x=\frac{3}{7}\)
c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=84\)
\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)
Vậy: x=30
d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)
hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)
Vậy: x=-5
e) Ta có: \(8\frac{2}{3}:x-10=-8\)
\(\Leftrightarrow\frac{26}{3}:x=2\)
hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)
Vậy: \(x=\frac{13}{3}\)
g) Ta có: \(x+30\%=-1.3\)
\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)
hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)
Vậy: \(x=\frac{-8}{5}\)
i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)
\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)
\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)
\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)
Vậy: x=-9
k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)
hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)
Vậy: x=30
m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)
\(\Leftrightarrow\left|2x-1\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)
\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=9\)
\(\Leftrightarrow19x+50=126\)
\(\Leftrightarrow19x=76\Leftrightarrow x=4\)
b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240
Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )
Tổng là : ( 30 + 1 ) . 30 : 2 = 465
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
Vậy........
( 4 x - 13 ) 4 + 4 3 = 145
( 4 x - 13 ) 4 + 64 = 145
( 4 x - 13 ) 4 = 81
( 4 x - 13 ) 4 = 3 4
=> 4 x - 13 = 3
4 x = 16
x = 4
3 x + 2 - 3 x = 72
3 x ( 3 2 - 1 ) = 72
3 x . 8 = 72
3 x = 9
3 x = 3 2
=> x = 2
2 x + 2 - 2 x - 1 = 224
2 x ( 2 2 - 2 -1 ) = 224
2 x . 3 , 5 = 224
2 x = 64
2 x = 2 6
=> x = 6
( 4 x - 13 ) 4 + 4 3 = 145
( 4 x - 13 ) 4 + 64 = 145
( 4 x - 13 ) 4 = 81
( 4 x - 13 ) 4 = 3 4
=> 4 x - 13 = 3
4 x = 16
x = 4
3 x + 2 - 3 x = 72
3 x ( 3 2 - 1 ) = 72
3 x . 8 = 72
3 x = 9
3 x = 3 2
=> x = 2
2 x + 2 - 2 x - 1 = 224
2 x ( 2 2 - 2 -1 ) = 224
2 x . 3 , 5 = 224
2 x = 64
2 x = 2 6
=> x = 6