Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x}{4}=\frac{y}{7}=\frac{2x}{8}=\frac{2x+y}{8+7}=\frac{45}{15}=3\)( ddo 22x2x + yy = 44545 )

\(\Rightarrow\hept{\begin{cases}x=3.4=12\\y=3.7=21\end{cases}}\)

\(\frac{x}{4}=\frac{y}{7}\Rightarrow y=\frac{7x}{4}\)

thay y = 7x/4 vào 2x+y=45 ta có

\(2x+\frac{7x}{4}=45\Rightarrow8x+7x=180\)

\(\Rightarrow15x=180\Rightarrow x=12\)

\(\Rightarrow y=\frac{7x}{4}=\frac{7\times12}{4}=21\)

\(\left(3x-2\right):45=3:\left(3x-2\right)\)

\(\Rightarrow3x-2=15\left(3x-2\right)\)

\(\Rightarrow3x-2=45x-30\)

\(\Rightarrow45x-3x=30-2\)

\(\Rightarrow42x=28\)

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

TL
 

x = \(\frac{2}{3}\) - \(\sqrt{15}\) 

x = \(\sqrt{15}\)+ \(\frac{2}{3}\)

HT

Câu 7 

\(A.x^{12}\)

Câu 8

\(D.\frac{7}{25}\)

TL ;

Câu 5 : B

Câu 6 : A

HT

Câu 5 : Chọn B

Lời giải :

\(\left(-\frac{3}{7}+\frac{3}{5}\right).\frac{4}{11}+\left(-\frac{4}{7}+\frac{2}{5}\right).\frac{4}{11}\)

\(=\frac{4}{11}\left(-\frac{3}{7}+\frac{3}{5}-\frac{4}{7}+\frac{2}{5}\right)\)

\(=\frac{4}{11}\left(-\frac{3}{7}-\frac{4}{7}+\frac{3}{5}+\frac{2}{5}\right)\)

\(=\frac{4}{11}\left(-1+1\right)\)

\(=\frac{4}{11}.0=0\)

TL ;

Câu 3 ; D

Câu 4 : A

HT

là c nha bạn

TL :

Câu 1 : Đáp án C

HT

\(\left[\left(-\frac{1}{5}\right)^4\right]^2=\left(-\frac{1}{5}\right)^{4.2}=\left(-\frac{1}{5}\right)^8=\left(\frac{1}{5}\right)^8\)

\(\left(-\frac{1}{5}\right)^8=\left(\frac{1}{5}\right)^8\)

Do vậy :

\(\left[\left(-\frac{1}{5}\right)^4\right]^2=\left(-\frac{1}{5}\right)^8\)

Vậy chọn B . 

0,(3) = \(\frac{1}{3}\)

Ta có :

 0,(3) = 0,333.....

\(\Rightarrow\)0,(3) = \(\frac{1}{3}\)

Vậy 0,(3)  = \(\frac{1}{3}\)