\(\frac{4}{10}:\left|x\right|=\left(\frac{6}{10}:\frac{3}{10}\right)^4\)

\(\frac{2}{5}:\left|x\right|=2^4\)

\(\left|x\right|=0,025\)

\(\Rightarrow\hept{\begin{cases}x=0,025\\x=-0,025\end{cases}}\)

\(\frac{4}{10}:\left|x\right|=\left(\frac{6}{10}\right)^4:\left(\frac{3}{10}\right)^4\)

\(\frac{4}{10}:\left|x\right|=2\)

\(\left|x\right|=\frac{4}{10}:2\)

\(\left|x\right|=\frac{1}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=-\frac{1}{5}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{5};-\frac{1}{5}\right\}\)

a)  \(P=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+...+\frac{9+10}{9^2.10^2}\)

\(P=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\) ( rút gọn số mũ nhé )

\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

\(P=1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

Vì \(\frac{9}{10}< 1\Rightarrow P< 1\) (đpcm)

b) Chút nữa mình làm nhé ^^

b) 

\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

Ta so sánh giữa A và Q.

\(\frac{1}{1.2}>\frac{1}{3};\frac{1}{2.3}>\frac{1}{3^2};\frac{1}{3.4}>\frac{1}{3^3};....;\frac{1}{100.101}>\frac{1}{3^{100}}\)

\(\Rightarrow Q< A\)

Ta lại tiếp tục so sánh A và \(\frac{1}{2}\)

Ta có :

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\Leftrightarrow A< \frac{1}{2}\)

Ta được:

\(Q< A< \frac{1}{2}\Leftrightarrow Q< \frac{1}{2}\)

\(\frac{x-3}{7-5x}=\frac{1}{x-2}\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=7-5x\)

\(\Rightarrow x^2-2x-3x+6=7-5x\)

\(\Rightarrow x^2-2x-3x+5x=7-6\)

\(\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

k nhé,Vy Nguyễn Đặng Khánh !

nhân tích chéo

\(\frac{x-3}{7-5x}=\frac{1}{x-2}\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=1\left(7-5x\right)\)

\(\Leftrightarrow x^2-3x-2x+6=7-5x\)

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow x^2=1\Leftrightarrow x=1\)

vậy x=1

Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta được:

    \(\frac{x+2}{x+6}=\frac{3}{x+1}\)

      \(\Rightarrow\left(x+1\right)\left(x+2\right)=3\left(x+6\right)\)

       \(\Leftrightarrow x^2+3x+2=3x+18\)

        \(\Leftrightarrow x^2=16\)

 Vậy \(x\in\left\{4;-4\right\}\)

(x+2)/(x+6)=3/(x+1)

<=>  (x+2)(x+1)/(x+6)(x+1)=3(x+6)/(x+6)(x+1)

=>(x+2)(x+1)=3(x+6)

<=> x^2+x+2x+2=3x+18

<=> x^2=16

<=>x^2=4^2 hoặc (-4)^2

<=> x=4 hoặc x=-4

Vậy......... 

Ta có:\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{2x}{6}=\frac{5y}{20}\)

   Áp dụng t/c dãy tỉ số bằng nhau ta đc:

       \(\Rightarrow\frac{2x}{6}=\frac{5y}{20}=\frac{2x+5y}{26}=\frac{5}{13}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=\frac{5}{13}\\\frac{y}{4}=\frac{5}{13}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{15}{13}\\y=\frac{20}{13}\end{cases}}\)

Ta có:

\(\frac{x}{3}=\frac{y}{4}\) và \(2x+5y=10\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x}{3}=\frac{y}{4}=\frac{2x+5y}{2.3+5.4}=\frac{10}{26}=\frac{5}{13}\)

\(\hept{\begin{cases}\frac{x}{3}=\frac{5}{13}\Rightarrow x=\frac{5}{13}.3=\frac{15}{13}\\\frac{y}{4}=\frac{5}{13}\Rightarrow y=\frac{5}{13}.4=\frac{20}{13}\end{cases}}\)

Vậy \(x=\frac{15}{13};y=\frac{20}{13}\)