Rút gọn biểu thức :
\(A=4\left(x-3\right)-3\left|x+3\right|\)
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4 tháng 11 2021
\(=\left(x-3\right)\left(x^2+1-x^2+1\right)=2\left(x-3\right)\)
CC
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TN
15 tháng 9 2018
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-3\right)^3-\left(x-2\right)^3\)
\(=x^3+2x^2+4x-2x^2-4x-8\) \(-\left(x^3-3x^2.3+3x.3^2-27\right)-\)-\(\left(x^3-3.x^2.2+3.x.2^2-8\right)\)
\(=x^3-8\) \(-x^3+9x^2-27x+27-x^3+6x^2-12x+8\)
\(=-x^3+15x^2-39x+27\)
học tốt
15 tháng 8 2023
\(A=log_2\left(x^3-x\right)-log_2\left(x+1\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x^3-x}{x+1}\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{x+1}\right)-log_2\left(x-1\right)\)
\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)=log_2x\)
A
19 tháng 6 2019
\(\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)\left(x^2+x+1\right).\)
\(=\left(x-1\right)^3+4\left(x+1\right)\left(1-x\right)+3\left(x-1\right)^3.\)
\(=\left(x-1\right)^3+4\left(1-x^2\right)+3\left(x-1\right)^3.\)
\(=\left(x-1\right)^3+3\left(x-1\right)^3+4\left(1-x^2\right)\)
\(=4\left(x-1\right)^3+4\left(1-x^2\right)\)
\(=4\left[\left(x-1\right)^3+\left(1-x^2\right)\right]\)
4 tháng 4 2021
Ta có: \(A=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}+\dfrac{2x+4}{4-x^2}\right)\cdot\left(x+\dfrac{5}{x-3}\right)\)
\(=\left(\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\left(\dfrac{x\left(x-3\right)+5}{\left(x-3\right)}\right)\)
\(=\dfrac{x^2-4x+4+x^2+2x-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x^2-4x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2-3x+5}{x-3}\)
\(=\dfrac{2x\left(x^2-3x+5\right)}{\left(x+2\right)\left(x-3\right)}\)
24 tháng 11 2021
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
\(A=4\left(x-3\right)-3\left|x+3\right|\)
- Nếu \(x\ge-3\) . Ta có : \(A=4.\left(x-3\right)-3.\left(x+3\right)=4x-12-3x-9=x-3\)
- Nếu \(x< -3\) . Ta có :
\(A=4.\left(x-3\right)-3.\left(-x-3\right)=4x-12+3x+9=x+21\)
\(A=4x-12-3\left|x+3\right|\)
(+) Với |x+3|=x+3
Thay vào biểu thưc ta được
\(A=4x-12-3\left(x+3\right)=4x-12-3x-9=x-21\)
(+) Với |x+3| = - (x+3)
Thay vào biểu thưc ta được
\(A=4x-12-3\left(-x-3\right)=4x-12+3x+9=7x-3\)