tìm x
a) 2x x 4 = 128
b) 2x(22)2 =(23)2
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DN
3
17 tháng 7 2021
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
17 tháng 7 2021
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
NA
0
CM
17 tháng 5 2017
a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.
b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.
c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.
d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.
NN
3
25 tháng 9 2021
a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
c) \(\Rightarrow2^x=32\Rightarrow x=5\)
d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)
e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)
f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)
HP
25 tháng 9 2021
a. 2x . 4 = 128
<=> 2x + 2 = 27
<=> x + 2 = 7
<=> x = 5
b. (2x + 1)3 = 125
<=> (2x + 1)3 - 53 = 0
<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)
<=> (2x - 4)(4x2 + 4x + 1 + 10x + 5 + 25) = 0
<=> (2x - 4)(4x2 + 14x + 31) = 0
<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)
c. 2x - 26 = 6
<=> 2x = 32
<=> x = 5
d. 64 . 4x = 45
<=> 43 . 4x = 45
<=> 43 + x = 45
<=> 3 + x = 5
<=> x = 2
e. 27 . 3x = 243
<=> 33 . 3x = 35
<=> 33 + x = 35
<=> 3 + x = 5
<=> x = 2
g. 49 . 7x = 2401 (Bn xem lại đề câu này)
<=> 72 . 7x = 74
<=> 72 + x = 74
<=> 2 + x = 4
<=> x = 2
23 tháng 9 2021
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
23 tháng 9 2021
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
a) 2x x 4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
b) 2x(22)2= (23)2
2x x 24 = 26
2x = 26 : 24
2x = 22
=> x = 2
a) 5
b) 2