`Answer:`

C D E F

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`

toan lop 6 nhe

`Answer:`

Theo đề ra, ta có: `A(x)=G(x)=>x^2 +1=x^2 +x+2`

`<=>x^2 -x^2 -x=(-1)+2`

`<=>-x=1`

`<=>x=-1`

này bạn hỏi vậy thế cho mk biết 2 kết quả bằng phải bằng số nào chứ ???

`Answer:`

\(\left|x-3\right|=\left|2x-1\right|\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=2x-1\\x-3=1-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2x=\left(-1\right)+3\\x+2x=1+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=2\\3x=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{3}\end{cases}}\)

`Answer:`

\(H=\frac{ab+b+2c}{b+c}+\frac{bc+c+2a}{c+a}+\frac{ac+a+2b}{a+b}\)

Ta có:

\(2c=2.\left(1-a-b\right)=2-2a-2b\)

\(\Rightarrow ab+b+2c=ab+b+2-2a-2b\)

\(=ab-2a-b+2\)

\(=a.\left(b-2\right)-\left(b-2\right)\)

\(=\left(b-2\right).\left(a-1\right)\)

\(=\left(b-2\right).\left(-b-c\right)\)

\(=-\left(b-2\right).\left(b+c\right)\)

\(\Rightarrow\frac{ab+b+2c}{b+c}=\frac{-\left(b-2\right).\left(b+c\right)}{b+c}=2-b\)

Chứng minh tương tự \(\Rightarrow\frac{bc+c+2a}{c+a}=2-c\)

                                     \(\Rightarrow\frac{ac+a+2b}{a+b}=2-a\)

\(\Rightarrow H=2-b+2-c+2-a\)

\(=6-\left(a+b+c\right)\)

\(=6-1\)

\(=5\)