1)

\(\sqrt{9\left(x-1\right)}=21\left(x\ge1\right)\\ < =>9\left(x-1\right)=441\\ < =>x-1=49\\ < =>x=50\left(tm\right)\)

2)

\(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\left(x\le1\right)\)

\(< =>\sqrt{1-x}+\sqrt{4\left(1-x\right)}-\dfrac{1}{3}\sqrt{16\left(1-x\right)}+5=0\\ < =>\sqrt{1-x}+2\sqrt{1-x}-\dfrac{1}{3}\cdot4\sqrt{1-x}=-5\\ < =>\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}=-5\\ < =>\dfrac{5}{3}\sqrt{1-x}=-5\\ < =>\sqrt{1-x}=-3\left(vl\right)\)

3)

\(\sqrt{2x}-\sqrt{50}=0\left(x\ge0\right)\\ < =>\sqrt{2x}=\sqrt{50}\\ < =>2x=50\\ < =>x=25\left(tm\right)\)

4)

\(\sqrt{4x^2+4x+1}=6\\ < =>\sqrt{\left(2x+1\right)^2}=6\\ < =>\left|2x+1\right|=6\\ < =>\left[{}\begin{matrix}2x+1=6\left(2x+1\ge0< =>x\ge-\dfrac{1}{2}\right)\\-2x-1=6\left(2x+1< 0< =>x< -\dfrac{1}{2}\right)\end{matrix}\right.\)

với `x>=-1/2`

`2x+1=6`

`<=>2x=5`

`<=>x=5/2(tm)`

với `x<-1/2`

`-2x-1=6`

`<=>-2x=7`

`<=>x=-7/2(tm)`

5)

\(\sqrt{\left(x-3\right)^2}=3-x\\ < =>\left|x-3\right|=3-x\\ < =>\left[{}\begin{matrix}x-3=3-x\left(x-3\ge0< =>x\ge3\right)\\3-x=3-x\left(x-3< 0< =>x< 3\right)\end{matrix}\right.\)

với `x>=3`

`x-3=3-x`

`<=>2x=6`

`<=>x=3`

với `x<3`

`3-x=3-x`

`<=>0x=0` (luôn đúng)

bạn tham khảo cái hình ảnh nhé bạn

`y=3x- 4`

với `x=0=>y=3*0-4=-4`

Vậy điểm `A(0;-4)` thuộc đồ thị hàm số

với `x=1=>y=3*1-4=-1`

Vậy điểm `B(1;-1)` thuộc đồ thị hàm số

Kẻ đường thằng AB để được đồ thị hàm số `y=3x-4`

`x xx 4-16=24`

`x xx 4=24+16`

`x xx 4=40`

`x=40:4`

`x=10`

phàn cuối đây ạ

`x(2x+7)=0`

`<=>x=0` hoặc `2x+7=0`

`<=>x=0` hoặc `x=-7/2`

`x(2x+7)>0`

\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\2x+7>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\2x+7< 0\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>-\dfrac{7}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< -\dfrac{7}{2}\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x>0\\x< -\dfrac{7}{2}\end{matrix}\right.\)

`x(2x+7)<0`

\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\2x+7< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\2x+7>0\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< -\dfrac{7}{2}\end{matrix}\right.\left(voli\right)}\\\left\{{}\begin{matrix}x< 0\\x>-\dfrac{7}{2}\end{matrix}\right.\end{matrix}\right.\\ < =>-\dfrac{7}{2}< x< 0\)

\(\sqrt{\dfrac{\left(1-\sqrt{5}\right)^2}{75}}\\ =\dfrac{\left|1-\sqrt{5}\right|}{\sqrt{75}}\\ =\dfrac{\sqrt{5}-1}{5\sqrt{3}}\left(vì1-\sqrt{5}< 0\right)\)

\(A=\dfrac{3\cdot\dfrac{a}{b}-\dfrac{-a}{b}}{-\dfrac{-5a}{b}+\dfrac{4a}{b}}\\ =\left(\dfrac{3a}{b}+\dfrac{a}{b}\right):\left(\dfrac{5a}{b}+\dfrac{4a}{b}\right)\\ =\dfrac{4a}{b}:\dfrac{9a}{b}\\ =\dfrac{4a}{b}\cdot\dfrac{b}{9a}\\ =\dfrac{4}{9}\)

Vậy `a=2021/2022` ; `b=2023/2022` thì `A=4/9`

`(1/2)^15 * (1/4)^20`

`=(1/2)^15 * [(1/2)^2]^20`

`=(1/2)^15 * (1/2)^40`

`=(1/2)^55`

`=1/(2^55)`