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\(a\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\\ =>\left(x-\frac{1}{2}\right)=\frac{1}{3}\\ =>x=\frac{1}{3}+\frac{1}{2}\\ =>x=\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\\ =>\left(x+\frac{1}{2}\right)=\frac{2}{5}\\ =>x=\frac{-1}{10}\)
d) (2x+3)2016=(2x+3)2018 khi 2x+3=0 hoặc 1
Nếu 2x+3=0
=2x=-3 ( loại )
Nếu 2x+3=1
=>2x=-2
=>x=-1 ( thỏa )

a,\(x^4:9=9\\ \Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
b,\(x^2-30=6\\ \Leftrightarrow x^2=36\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
c,\(x^2:10=10\Leftrightarrow x^2=\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
x4 : 9 = 9
x4 = 9 . 9
x4 = 81
x4 = 34
=> x = 3
x2 - 30 = 6
x2 = 6 + 30
x2 = 36
x2 = 62
=> x = 6
x2 : 10 = 10
x2 = 10 . 10
x2 = 100
x2 = 102
=> x = 10

Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0

a: \(x\in B\left(5\right)\)
mà x<10
nên \(x\in\left\{...;0;5\right\}\)
b: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
c: \(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{3;1;7;-3\right\}\)
d: \(\Leftrightarrow x+1\in\left\{1;-1;10;-10;2;-2;5;-5\right\}\)
hay \(x\in\left\{0;-2;9;-11;1;-3;4;-6\right\}\)
e: \(\Leftrightarrow x^2+1\in\left\{1;2;5;10\right\}\)
hay \(x\in\left\{0;1;-1;3;-3\right\}\)

a) \(2^3.2^x=64\Leftrightarrow8.2^x=64\Leftrightarrow2^x=\dfrac{64}{8}=8=2^3\Rightarrow x=3\)
vậy \(x=3\)
b) \(7.7^x=343\Leftrightarrow7^x=\dfrac{343}{7}=49=7^2\Rightarrow x=2\)
vậy \(x=2\)
c) \(7.7^{x+1}=343\Leftrightarrow7^{x+1}=\dfrac{343}{7}=49=7^2\Rightarrow x+1=2\Leftrightarrow x=1\)
vậy \(x=1\)
d) \(2^x-15=17\Leftrightarrow2^x=17+15=32=2^5\Rightarrow x=5\)
vậy \(x=5\)
e) \(x^{10}=1\Leftrightarrow x^{10}=1^{10}\Rightarrow x=1\)
vậy \(x=1\)
\(x^{10}=x\Leftrightarrow x^{10}-x=0\Leftrightarrow x\left(x^9-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy \(x=0;x=1\)
a) \(2^3.2^x=2^3.2^3\)
=> \(2^x=2^3\)
=> x=3
b) \(7.7^x=343\)
=> \(7.7^x=7.7^2\)
=> \(7^x=7^2\)
=> x=2
c) \(7.7^{x+1}=7.7^2\)
=> \(7^{x+1}=7^2\)
=> x+1=2
=> x=1
d) \(2^x-15=17\)
=> \(2^x=17+15\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> x=5
e) \(x^{10}=1\)
=> \(x^{10}=1^{10}\)
=> x=1
g) \(x^{10}=x\)
=> x = 0 hoặc 1
+) Với x=0 => \(0^{10}=0\)( t/m )
+) Với x=1 => \(1^{10}=1\)( t/m )

\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=140\)
\(\Leftrightarrow\left(4x+720\right):x=140\)
\(\Leftrightarrow4x+720=140.x\)
\(\Leftrightarrow4x-140x=-720\)
\(\Leftrightarrow x.\left(-136\right)=-720\)
\(\Leftrightarrow x=-720:\left(-136\right)\)
\(\Leftrightarrow x=\frac{90}{17}\)
\(2\)) Mình đang nghĩ
5
\(x=10.x:2.x\)
\(x\) = (10 : 2).\(x.x\)
\(x\) = 5.\(x^2\)
\(5x^2-x=0\)
\(x\left(5x-1\right)=0\)
\(\left[\begin{array}{l}x=0\\ 5x-1=0\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ 5x=1\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=\frac15\end{array}\right.\)
Vậy \(x\in\) {0; \(\frac15\)}