Xin lỗi bạn mình mới học lớp 5 thôi

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còn mình mới học lớp 4

Ta có :\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}+\frac{12}{x^2+8}+\frac{7}{x^2+3}=3\)

\(\Leftrightarrow\left(\frac{6}{x^2+2}-1\right)+\left(\frac{12}{x^2+8}-1\right)+\left(\frac{7}{x^2+3}-1\right)=0\)

\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+3}=0\)

\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

Ta thấy : \(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\forall x\)

Do đó : \(4-x^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\) ( thỏa mãn )

Vậy : \(x\in\left\{-2,2\right\}\)

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> \(4\left(x+1\right)=3\left(x-2\right)\)

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

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