2.f(x) - x.f(-x) = x + 10 (1) 
Thay x = -x ta được 2f(-x) + xf(x) = -x + 10 (2) 
(1)*2 + (2)*x  =>  4f(x) + x^2f(x) = 2x + 20 - x^2 + 10x 
=>  f(x) = (-x^2 + 12x + 20)/(x^2 + 4) 
=>  f(2) = 5

Ta có: \(\left(\left(\frac{2}{3}\right)^x\right)^3=\left(\frac{3}{2}\right)^3\)

=>   \(\left(\frac{2}{3}\right)^x=\frac{3}{2}\)

=>   \(\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{-1}\)

=>   x =  -1

\(\left(\frac{2}{3}^x\right)^3=\frac{27}{8}\)

\(\left(\frac{2}{3}^x\right)^3=\left(\frac{3}{2}\right)^3\)

=> \(\frac{2}{3}^x=\frac{3}{2}\)

\(\frac{2}{3}^x=\frac{2}{3}^{-1}\)

=> \(x=-1\)

• Viết \(\frac{-4}{5}=\frac{-4\cdot9}{9\cdot9}=\frac{9}{\frac{-81}{4}}=\frac{9}{-20\frac{1}{4}}.\)và \(\frac{-3}{5}=\frac{-3\cdot\left(-3\right)}{5\cdot\left(-3\right)}=\frac{9}{-15}\)
• Vậy có 5 phân số TM bài toán là: \(\frac{9}{-20};\frac{9}{-19};\frac{9}{-18};\frac{9}{-17};\frac{9}{-16}\)

a)

Có \(\frac{4}{5}< 1\)  ; \(1,1>1\)

=> \(\frac{4}{5}< 1,1\)

b)

Có \(-500< 0\) ; \(0,001>0\)

 => \(-500< 0,001\)

c)      Có 

\(\frac{13}{38}=1-\frac{25}{38}\) ; \(\frac{-12}{-37}=\frac{12}{37}=1-\frac{25}{37}\)

Vì \(\frac{25}{38}< \frac{25}{37}=>1-\frac{25}{38}>1-\frac{25}{37}\)

=> \(\frac{13}{38}>\frac{12}{37}=>\frac{13}{38}>\frac{-12}{-37}\)

Ta có \(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5.\)

\(\Rightarrow\frac{2}{3}-\frac{1}{3}.x+\frac{1}{3}.\frac{3}{2}-\frac{1}{2}.2x-\frac{1}{2}=5\)

\(\Rightarrow\frac{2}{3}-\frac{x}{3}+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\Rightarrow\frac{4}{6}-\frac{2x}{6}+\frac{3}{6}-\frac{6x}{6}-\frac{3}{6}=\frac{30}{6}\)

\(\Rightarrow4-2x+3-6x-3=30\)

\(\Rightarrow4-8x=30\)

\(\Rightarrow-8x=26\)

\(\Rightarrow x=\frac{26}{-8}=-\frac{13}{4}\)

Vậy \(x=-\frac{13}{4}\)

Kết quả góc = 80 nhé. 

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kết quả là 80

k mk nha