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\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
làm bài & thôi :
(x2 - 2x + 3) \(⋮\)(x - 1)
= x2 - 2x + 3
=) x2 - 2x + 3 - ( x - 1 )
=) x2 - 1
=) x2 - 1 - x( x - 1 )
=) 2 \(⋮\)x - 1
tự làm
a) Ta có: (x2 - 2x + 3) \(⋮\)(x - 1)
<=> [x(x - 1) - (x - 1) + 2] \(⋮\)(x - 1)
<=> [(x - 1)2 + 2] \(⋮\)(x - 1)
Do (x - 1)2 \(⋮\)(x - 1) => 2 \(⋮\)(x - 1)
=> (x - 1) \(\in\)Ư(2) = {1; -1; 2; -2}
Lập bảng :
| x - 1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
Vậy ...
b) (3x - 1) \(⋮\)(x - 4)
<=> [3(x - 4) + 11] \(⋮\)(x - 4)
Do 3(x - 4) \(⋮\)(x - 4) => 11 \(⋮\)(x - 4)
=> (x - 4) \(\in\)Ư(11) = {1; -1; 11; -11}
Lập bảng:
| x - 4 | 1 | -1 | 11 | -11 |
| x | 5 | 3 | 15 | -7 |
vậy ...
c;d tương tự trên
\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)
\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)
\(2^x\cdot\frac{9}{2}=288\)
\(2^x=64\)
\(2^x=2^6\)
\(x=6\)
\(9^x:3^x=3^7\)
\(3^{2x}:3^x=3^7\)
\(3^x=3^7\)
\(x=7\)
\(7^{x+2}+2\cdot7^{x-1}=345\)
\(7^x\cdot7^2+2\cdot7^x:7=345\)
\(7^x\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\cdot\frac{345}{7}=345\)
\(7^x=7\)
\(x=1\)
a) 1/2.2^x + 2^x+2 = 256 + 32
1/2.2^x + 2^2.2^x=288
2^x(1/2+4)= 288
2^x.4,5=288
2^x= 288:4,5
2^x=64=2^6
x=6
\(x+1⋮x-1\)
\(x-1+2⋮x-1\)
\(2⋮x-1\)hay \(x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x - 1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
\(x-2⋮x+1\)
\(x+1-3⋮x+1\)
\(-3⋮x+1\)thự hiện tương tự nhé !
\(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}-\frac{1}{3}+1\)
\(\Rightarrow-7x=\frac{-1}{6}\)
\(\Rightarrow x=\frac{1}{42}\)
Vậy ...
\(\)
\(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{18}\)
Vậy...
a) -5/2 - 1/2x = -19/5
1/2x=-5/2 - (-19/5)
1/2x=13/10
x=13/10:1/2
x=13/5