Ta có: \(3\left|x^2-1\right|-6=\left|1-x^2\right|\)

\(\Leftrightarrow3\left|x^2-1\right|-\left|x^2-1\right|=6\)

\(\Leftrightarrow2\left|x^2-1\right|=6\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=3\\x^2-1=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=-2\end{cases}}\)

Vì \(x\ge0>-2\left(\forall x\right)\)

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

Vậy \(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

là câu A

tin mình đi

\(\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)

Để phân số có giá trị nguyên => \(\frac{3}{n-2}\)có giá trị nguyên

<=> \(3⋮n-2\)=> \(n-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=\frac{3}{n-2}\)

\(\Rightarrow n-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Ta lập bảng : 

Tỉ số % của 20 và 80 là:

\(\frac{20}{80}\times100=25\%\)

=> (B) đúng

Tỉ số % của 20 và 80 là

 A 250%,

B 25%

C 2,5%

D 0,25%

\(\frac{3}{4}=\frac{x}{-4}=\frac{21}{y}=\frac{z}{80}\)

Vì \(\frac{3}{4}=\frac{x}{-4}\)\(\Rightarrow\frac{3}{4}=\frac{-x}{4}\)\(\Rightarrow-x=3\)\(\Rightarrow x=-3\)

Vì \(\frac{3}{4}=\frac{21}{y}\)\(\Rightarrow y=\frac{21.4}{3}=28\)

Vì \(\frac{3}{4}=\frac{z}{80}\)\(\Rightarrow z=80.\frac{3}{4}=60\)

Vậy \(x=-3\); \(y=28\); \(z=60\)

$1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+...+20\right)$

$1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2$

$=\frac{2}{2}+\frac{3}{2}+...+\frac{21}{2}$

$=\frac{2+3+...+21}{2}$

$=\frac{230}{2}$

$=115$

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}+\left(1+2+3\right)+...+\frac{1}{20}+\left(1+2+...+20\right)\)

= \(1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2\)

= \(\frac{2}{2}+\frac{3}{2}+...+\frac{21}{2}\)

=\(\frac{2+3+...+21}{2}\)

= \(\frac{230}{2}=115\)

Học tốt!

\(6x-7,2x=-14,4\)

\(\Leftrightarrow-1,2x=-14,4\)

\(\Leftrightarrow x=12\)

Vậy x = 12

x.(6-7,2)=-14,4

x.-1,2=-14,

4x=-14,4:-1,2

x=12

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Sửa đề : \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{42}{43}\)