a) \( - \left( {4 + 7} \right) =  - 11\)

\(\begin{array}{l}\left( { - 4 - 7} \right) = \left( { - 4} \right) + \left( { - 7} \right)\\ =  - \left( {4 + 7} \right) =  - 11\\ \Rightarrow \left( { - 4 - 7} \right) =  - \left( {4 + 7} \right)\end{array}\)

b)

\(\begin{array}{l} - \left( {12 - 25} \right) =  - \left[ {12 + \left( { - 25} \right)} \right]\\ =  - \left[ { - \left( {25 - 12} \right)} \right] =  - \left( { - 13} \right) = 13\end{array}\)

\(\begin{array}{l}\left( { - 12 + 25} \right) = 25 - 12 = 13\\ \Rightarrow  - \left( {12 - 25} \right) = \left( { - 12 + 25} \right)\end{array}\)

c)

\(\begin{array}{l} - \left( { - 8 + 7} \right) =  - \left[ { - \left( {8 - 7} \right)} \right] =  - \left( { - 1} \right) = 1\\\left( {8 - 7} \right) = 1\\ \Rightarrow  - \left( { - 8 + 7} \right) = \left( {8 - 7} \right)\end{array}\)

d)

\(\begin{array}{l} + \left( { - 15 - 4} \right) =  + \left[ {\left( { - 15} \right) + \left( { - 4} \right)} \right]\\ =  + \left[ { - \left( {15 + 4} \right)} \right] =  + \left( { - 19} \right) =  - 19\\\left( { - 15 - 4} \right) = \left( { - 15} \right) + \left( { - 4} \right)\\ =  - \left( {15 + 4} \right) =  - 19\\ \Rightarrow  + \left( { - 15 - 4} \right) = \left( { - 15 - 4} \right)\end{array}\)

e)

\(\begin{array}{l} + \left( {23 - 12} \right) =  + 11 = 11\\\left( {23 - 12} \right) = 11\\ \Rightarrow  + \left( {23 - 12} \right) = \left( {23 - 12} \right)\end{array}\)

\(\left( { - 1} \right) + \left( { - 3} \right) =  - \left( {1 + 3} \right) =  - 4\)

\(\left( { - 3} \right) + \left( { - 1} \right) =  - \left( {3 + 1} \right) =  - 4\)

\( \Rightarrow \left( { - 1} \right) + \left( { - 3} \right) = \left( { - 3} \right) + \left( { - 1} \right)\)

\(\left( { - 7} \right) + \left( { + 6} \right) =  - \left( {7 - 6} \right) =  - 1\)

\(\left( { + 6} \right) + \left( { - 7} \right) =  - \left( {7 - 6} \right) =  - 1\)

\( \Rightarrow \left( { - 7} \right) + \left( { + 6} \right) = \left( { + 6} \right) + \left( { - 7} \right)\)

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

\(a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4\)

\(=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}\)

\(b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)\)

\(=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}\)

\(=\frac{100.12}{123.16}+\frac{23.15}{123.20}\)

\(=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}\)

\(=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}\)

\(=\frac{123}{164}=\frac{3}{4}\)

a, bạn tự làm 

b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)

\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)

c, đk : \(x\ne\dfrac{2}{3}\)

Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)

Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)

Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)

a: Vì 0,2<1

nên hàm số \(y=\left(0,2\right)^x\) nghịch biến trên R

mà -3<-2

nên \(\left(0,2\right)^{-3}>\left(0,2\right)^{-2}\)

b: Vì \(0< \dfrac{1}{3}< 1\)

nên hàm số \(y=\left(\dfrac{1}{3}\right)^x\) nghịch biến trên R

mà \(2000< 2004\)

nên \(\left(\dfrac{1}{3}\right)^{2000}>\left(\dfrac{1}{3}\right)^{2004}\)

c: Vì 3,2>1

nên hàm số \(y=\left(3,2\right)^x\) đồng biến trên R

mà \(1,5< 1,6\)

nên \(\left(3,2\right)^{1,5}< \left(3,2\right)^{1,6}\)

d: Vì \(0< 0,5< 1\)

nên hàm số \(y=\left(0,5\right)^x\) nghịch biến trên R

mà -2021>-2023

nên \(\left(0,5\right)^{-2021}< \left(0,5\right)^{-2023}\)

a: 78x(x-97)-x+97=0

=>(x-97)(78x-1)=0

=>\(\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

b: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

=>\(x\left(x-2\right)\left(x+2\right)=0\)

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

c: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

=>\(x^2+4x+4-x^2+4=0\)

=>4x+8=0

=>x+2=0

=>x=-2

\(a,78x\left(x-97\right)-x+97=0\)

\(\Leftrightarrow78x\left(x-97\right)-\left(x-97\right)=0\)

\(\Leftrightarrow\left(x-97\right)\left(78x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-97=0\\78x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

\(b,\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

\(c,\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\cdot4=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)