cứu em với mn ơi 

cái này hỏi trời nhá em, tự làm nha, không lươiif

8: Ta có: \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\)

\(=\sqrt{5}+1-\sqrt{5}+1\)

=2

Câu 1.

Khi mở khóa K:

\(I_m=I_1=0,4A\)

Khi đóng khóa K:

\(I_m=I_1+I_2=0,6\Rightarrow I_2=0,2A\)

\(U_1=0,4\cdot5=2V\)

\(\Rightarrow U_2=U_1=2V\)

\(\Rightarrow U=U_1=U_2=2V\)

\(R_2=\dfrac{U_2}{I_2}=\dfrac{2}{0,2}=10\Omega\)

Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

Câu 6: B

Câu 7: A

one - by- of - like - saying - no - made - about - who - never

-Bài 3:

2) -Áp dụng BĐT Caushy Schwarz ta có:

\(A=\dfrac{1}{x^3+3xy^2}+\dfrac{1}{y^3+3x^2y}\ge\dfrac{\left(1+1\right)^2}{x^3+3xy^2+3x^2y+y^3}=\dfrac{4}{\left(x+y\right)^3}\ge\dfrac{4}{1^3}=4\)-Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

a) Ta có: \(\left(3x+1\right)^3\)

\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)

\(=27x^3+27x^2+9x+1\)

b) Ta có: \(\left(2x-\dfrac{1}{x}\right)^3\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot\dfrac{1}{x}+3\cdot2x\cdot\left(\dfrac{1}{x}\right)^2-\left(\dfrac{1}{x}\right)^3\)

\(=8x^3-12x+\dfrac{6}{x}-\dfrac{1}{x^3}\)

Chữ đẹp quá bạn ơi, không hiểu gì hết 

Câu 59:B

Câu 63: A