\(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Leftrightarrow\left(x+4\right)^2=100\)

\(\Rightarrow\orbr{\begin{cases}x+4=\sqrt{100}\\x+4=-\sqrt{100}\end{cases}}\Rightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)

\(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Rightarrow\left(x+4\right)^2=5\times20\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow\left(x+4\right)^2=\pm10\)

\(\Rightarrow\orbr{\begin{cases}x+4=10\\x+4=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10-4\\x=\left(-10\right)-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-14\end{cases}}\)

\(\left|x-1\right|\ge0\forall x\in R\)

\(\left(x-3\right)^4\ge0\forall x\in R\)

\(\Rightarrow\left(x-3\right)^4+\left|x-1\right|>0\)

Vậy đa thức D(x) vô nghiệm

\(A\left(x\right)=2x^2-4x+3+4x^3-6=4x^3+2x^2-4x-3\)

\(B\left(x\right)=-4x^3-4x+2x^2-x-3=-4x^3+2x^2-5x-3\)

a) \(A\left(x\right)+B\left(x\right)=4x^3+2x^2-4x-3+\left(-4x^3\right)+2x^2-5x-3\)

\(=\left(4x^3-4x^3\right)+\left(2x^2+2x^2\right)+\left(-4x-5x\right)+\left(-3-3\right)\\ =4x^2-9x-6\)

b) \(A\left(x\right)-B\left(x\right)=4x^3+2x^2-4x-3+4x^3-2x^2+5x+3\)

\(=\left(4x^3+4x^3\right)+\left(2x^2-2x^2\right)+\left(-4x+5x\right)+\left(-3+3\right)\\ =8x^3+x\)

\(a)A\left(x\right)+B\left(x\right)=\left(2x^2-4x+3+4x^3-6\right)+\left(-4x^3-4x+2x^2-x-3\right)\)

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

                                   \(=\left(2x^2+2x^2\right)+\left(-4x-4x-x\right)+\left(3-6-3\right)+\left(4x^3-4x^3\right)\)

                                   \(=4x^2+\left(-9x\right)+\left(-6\right)\)

                                   \(=4x^2-9x-6\)

\(b)A\left(x\right)-B\left(x\right)=\left(2x^2-4x+3+4x^3-6\right)-\left(-4x^3-4x+2x^2-x-3\right)\)

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

                                   \(=\left(2x^2-2x^2\right)+\left(-4x+4x+x\right)+\left(3-6+3\right)+\left(4x^3+4x^3\right)\)

                                  \(=x+\left(-6\right)+8x^3\)

                                  \(=x-6+8x^3\)

`Answer:`

Xét `\triangleCFD` và `\triangleCFE:`

`CF` chung

`CD=CE`

`FD=FE`

`=>\triangleCFD=\triangleCFE(c.c.c)`

`=>\hat{CFD}=\hat{CFE}` mà `\hat{CFD}+\hat{CFE}=180^o` (Kề bù) `=>\hat{CFD}=\hat{CFE}=90^o`

`=>CF` vuông góc `DE`

Áp dụng định lý Pytago vào `\triangleCFD` vuông tại `F:`

`CF^2+DF^2=CD^2 <=>24^2 +DF^2 =25^2 <=>576+DF^2 =625<=>DF^2=49<=>DF=7cm`

`=>DE=2DF=14cm`

Hok tốt:3~!

A=-112

`Answer:`

\(A=\left(-\frac{1}{2}x^2y\right)\left(\frac{14}{5}xy^3z^3\right)\)

\(=\left(-\frac{1}{2}.\frac{14}{5}\right).\left(x^2.x\right).\left(y.y^3\right).z^3\)

\(=-\frac{7}{5}x^3y^4z^3\)

Bậc `10`

Thay `x=-1` và `y=2` và `z=-1` vào đơn thức `A`, ta được:

\(A=-\frac{7}{5}.\left(-1\right)^3.2^4.\left(-1\right)^3=-\frac{7}{5}.-1.16.-1=-\frac{112}{5}\)

`Answer:`

\(\frac{x+2}{5}=\frac{2-3x}{3}\)

\(\Leftrightarrow\frac{3.\left(x+2\right)}{5}=\frac{5.\left(2-3x\right)}{3}\)

\(\Rightarrow3.\left(x+2\right)=5.\left(2-3x\right)\)

\(\Leftrightarrow3x+6=10-15x\)

\(\Leftrightarrow3x+15x=10-6\)

\(\Leftrightarrow18x=4\)

\(\Leftrightarrow x=\frac{2}{9}\)

\(\frac{x+2}{5}=\frac{2-3x}{3}\)

\(\Rightarrow3\times\left(x+2\right)=5\times\left(2-3x\right)\)

\(\Rightarrow3x+6=10-15x\)

\(\Rightarrow3x+15x=-6+10\)

\(\Rightarrow18x=4\)

\(\Rightarrow x=4\div18=\frac{2}{9}\)

`Answer:`

\(A=4+2^2+2^3+...+2^{99}\)

\(=2^2+2^2+2^3+...+2^{99}\)

Ta thấy:

\(2^2+2^2=2^2.2=2^3\)

\(2^3+2^3=2^3.2=2^4\)

...

\(2^{99}+2^{99}=2^{99}.2=100\)

\(\Rightarrow A=2^2+2^2+2^3+...+2^{99}=2^{100}\)

Mà \(2^{100}⋮2^{99}\)

`=>A` chia hết cho `2^99`