\(\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)-\left(1+\dfrac{a+\sqrt{a}}{1+\sqrt{a}}\right)\\ =\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)-\left(1+\dfrac{\sqrt{a}\left(1+\sqrt{a}\right)}{1+\sqrt{a}}\right)\\ =\left(1-\sqrt{a}\right)-\left(1+\sqrt{a}\right)\\ =1-a\left(đpcm\right)\)

Sửa lại đề nhé !

\(\left|x+\dfrac{1}{7}\right|-\dfrac{2}{3}=0\)

\(\Rightarrow\left|x+\dfrac{1}{7}\right|=0+\dfrac{2}{3}\\ \Rightarrow\left|x+\dfrac{1}{7}\right|=\dfrac{2}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{7}=\dfrac{2}{3}\\x+\dfrac{1}{7}=-\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}-\dfrac{1}{7}\\x=-\dfrac{2}{3}-\dfrac{1}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{21}\\x=-\dfrac{17}{21}\end{matrix}\right.\)

`(2^x+1)^2 =25`

`=> (2^x+1)^2 = (+-5)^2`

\(\Rightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

\(\left(x+6\right)\left(5^x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

\(\left(x-3\right)^{2023}=x-3\)

\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

`2x -2/3 +1/2x =-1`

`=> 2x+1/2x =-1+2/3`

`=> (2+1/2) x =-3/3 + 2/3`

`= 5/2 x = -1/3`

`=> x=-1/3 : 5/2`

`=>x= -2/15`

`31/36 - (1/3-x)^2 =5/6`

`=>  (1/3-x)^2 = 31/36 - 5/6`

`=>  (1/3-x)^2 =1/36`

`=>  (1/3-x)^2 = (+-1/6)^2`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{1}{6}\\\dfrac{1}{3}-x=-\dfrac{1}{6}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{1}{2}\end{matrix}\right.\)

`1/2 x -2 =2x+1`

`=> 1/2 x -2x = 1 +2`

`=> (1/2-2)x = 3`

`=> -3/2 x =3`

`=> x= 3 : (-3/2)`

`=>x=-2`

`a, 3x -27 =4*3^2`

`=> 3x -27 =4*9`

`=> 3x -27 = 36`

`=> 3x= 36+27`

`=> 3x=63`

`=> x= 63:3`

`=>x= 21`

`b, 62-(x+22)=2^3 *5`

`=> 62-(x+22)=8*5`

`=> 62-(x+22)=40`

`=> x+22= 62-40`

`=> x+22= 22`

`=> x=22-22`

`=>x=0`

`c, 4*(3x^3 +1^(10))=4*5^2`

`=> 4*(3x^3 +1) = 4*25`

`=> 4*(3x^3 +1) = 100`

`=> 3x^3 +1=100:4`

`=> 3x^3 +1=25`

`=> 3x^3 = 25-1`

`=> 3x^3 =24`

`=>x^3=24/3 =8`

`=> x^3 =2^3`

`=>x=2`

`x:2=y:(-5) => x/2=y/(-5)`

ADTC dãy tỉ số b/nhau ta có:

`x/2=y/(-5) =(x-y)/( 2+5) =-7/7=-1`

`=>x/2=-1=> x=-2`

`=> y/(-5)=-1=>y=5`

`A=100-99+98-97+96-95+...+2-1`

`=(100-99)+(98-97)+(96-95)+...+(2-1)`

`= 1 +1+1+....+1` ( có `50` cặp )

`= 1 xx 50`

`=50`