Question

Giúp em bài này vs ạ

Answer 1

\(a,3x^2-4x-10=\left(8x^2-5x+6\right)-\left(5x^2+2x-3\right)\)

\(3x^2-4x-10=8x^2-5x+6-5x^2-2x+3\)

\(3x^2-4x-10=3x^2-7x+9\)

\(3x=19\)

\(x=\frac{19}{3}\left(TM\right)\)

\(b,2\left(x^2-x\right)+\left(6x^2-4x+8\right)=6x+8x^2-1\)

\(2x^2-2x+6x^2-4x+8=6x+8x^2-1\)

\(8x^2-6x+8=8x^2+6x-1\)

\(-12x=-9\)

\(x=\frac{3}{4}\left(TM\right)\)

\(e,\left(x+2021\right)^{x+9}=\left(x+2021\right)^{x+7}\)

\(\left(x+2021\right)^{x+9}:\left(x+2021\right)^{x+7}=1\)

\(\left(x+2021\right)^2=1\)

\(x=-2020\left(TM\right)\)

\(f,\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)

\(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{2.3}\right|\ge0\\...\left|x+\frac{1}{99.100}\right|\ge0\end{cases}< =>\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge}0\)

\(< =>100x\ge0\)

\(x\ge0\)

\(< =>\left|x+\frac{1}{1.2}\right|=x+\frac{1}{1.2}\)

\(\left|x+\frac{1}{2.3}\right|=x+\frac{1}{2.3}\)... tương tự các cái còn lại

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)

\(x+\frac{1}{1.2}+x+\frac{1}{2.3}+....+x+\frac{1}{99.100}=100x\)

\(99x+\left(1-\frac{1}{2}+\frac{1}{2}-..........+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(99x+\left(1-\frac{1}{100}\right)=100x\)

\(x=\frac{99}{100}\left(TM\right)\)