nếu muốn làm nhanh thì dùng chức năng table trong máy tính cầm tay

Còn đây là cách giải:

\(\left(-x-4\right)^2-2\cdot\left|4+x\right|=0\)

\(\Leftrightarrow\left(-x\right)^2+8x+16-2\cdot\left|4+x\right|=0\)

\(\Leftrightarrow x^2+8x+16-2\cdot\left|4+x\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+8x+16-2\left(4+x\right)=0\left(đk:4+x\ge0\right)\\x^2+8x+16-2\cdot\left(-\left(4+x\right)\right)=0\left(đk:4+x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(đk:x\ge-4\right)\\x=-4\left(đk:x\ge-4\right)\\x=-4\left(đk:x< -4\right)\\x=-6\left(đk:x< -4\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=-6\end{matrix}\right.\)

Vậy \(x_1=-6;x_2=-4;x_3=-2\)

Xem giá bìa là 100% thì phải trả là :

100% - 10% = 90% (giá)

Giá bìa của quyển sách là :

7290 : 90% = 8100 (đồng)

       ĐS : 8100 đồng

Tick tớ nha   do thi ngoc linh

\(x^2-5x+3=0\)

\(\Leftrightarrow x=\dfrac{-\left(-5\right)\pm\sqrt{\left(-5\right)^2-4\cdot1\cdot3}}{2\cdot1}\)

\(\Leftrightarrow x=\dfrac{5\pm\sqrt{25-12}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{13}}{2}\\x=\dfrac{5-\sqrt{13}}{2}\end{matrix}\right.\)

a) \(ĐKXĐ:x\ne10;x\ne-10;x\ne0;x\ne-4\)

a) \(0,\left(27\right)+0,\left(72\right)=0,\left(99\right)=1\)

b) \(0,\left(22\right)\cdot\dfrac{9}{2}=\dfrac{2}{9}\cdot\dfrac{9}{2}=1\)

c) \(\left[0,\left(11\right)\cdot9\right]^{2003}=\left(\dfrac{1}{9}\cdot9\right)^{2003}=1^{2003}=1\)

a) \(0,\left(27\right)+0,\left(72\right)=0,\left(100\right)=1\)

b) \(0,\left(22\right)\cdot\dfrac{9}{2}=\dfrac{2}{9}\cdot\dfrac{9}{2}=1\)

c) \(\left[0,\left(11\right)\cdot9\right]^{2003}=\left(\dfrac{1}{9}\cdot9\right)^{2003}=1^{2003}=1\)

\(0,5=\dfrac{1}{2}\)

\(1,\left(45\right)=\dfrac{16}{11}\)

\(0,2\left(13\right)=\dfrac{211}{990}\)

\(0,04\left(3\right)=\dfrac{13}{300}\)

c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}\)

\(=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)\)

\(=-\left(\dfrac{x^2}{5}+\dfrac{y^4}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)

a) \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right)\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{x-1}{\sqrt{x}}\)

\(=\left(\dfrac{2\cdot2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\left(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+4\sqrt{x}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=4\sqrt{x}\cdot\left[1+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]\cdot\dfrac{1}{\sqrt{x}}\)

\(=4\left(1+x-1\right)\)

\(=4x\)

b) Thay \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) vào biểu thức A.

Ta có:

\(4\cdot\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}\cdot\left(2-\sqrt{6}\right)}{-2}\\ =-2\sqrt{6}\cdot\left(2-\sqrt{6}\right)=-4\sqrt{6}+12\)

Vậy giá trị biểu thức A tại \(x=\dfrac{\sqrt{6}}{2+\sqrt{6}}\) là \(-4\sqrt{6}+12\)

c) Để \(\sqrt{A}>A\)

\(\Rightarrow\sqrt{4x}>4x\)

\(\Leftrightarrow\sqrt{4x}>4x\left(đk:x\ge0\right)\)