3.4x^2+8.2x=0

<=>x(3.4x+8.2)=0

<=>x=0 hoặc 3.4x+8.2=0

<=>x=0 hoặc x=-8.2/3.4=-41/17

         3,4x² + 8,2x = 0

<=>   x(3,4x + 8,2) = 0

<=> \(\hept{\begin{cases}x=0\\3,4x+8,2=0\end{cases}\hept{\begin{cases}x=0\\x\approx2,41\end{cases}}}\)

Ta có:\(\hept{\begin{cases}x^2+y^2=5\left(1\right)\\xy=2\left(2\right)\end{cases}}\)

Từ (1) ta có:\(\left(x+y\right)^2-2xy=5\)

thay (2) vaò (1) \(\Rightarrow\left(x+y\right)^2=9\Leftrightarrow\orbr{\begin{cases}x+y=3\left(3\right)\\x+y=-3\left(4\right)\end{cases}}\)

Từ (2) \(\Rightarrow y=\frac{2}{x}\)Thay vào (3) ta có \(x+\frac{2}{x}=3\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=2\\x=2\Rightarrow y=1\end{cases}}\)

Cái kia tg tự nha bn

Vì \(a;b;c>0\Rightarrow2ab\le\frac{\left(a+b\right)^2}{2}\) thay vào \(\sqrt{a^2+4ab+b^2}\)ta có:

\(\sqrt{a^2+4ab+b^2}=\sqrt{\left(a+b\right)^2+2ab}\)

\(\le\sqrt{\left(a+b\right)^2+\frac{\left(a+b\right)^2}{2}}=\sqrt{\frac{3\left(a+b\right)^2}{2}}=\left(a+b\right).\sqrt{\frac{3}{2}}\)

Tương tự: \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}.\left(b+c\right)\)

\(\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}.\left(c+a\right)\)

\(\Rightarrow P\le\sqrt{\frac{3}{2}}.\left(a+b\right)+\sqrt{\frac{3}{2}}.\left(b+c\right)+\sqrt{\frac{3}{2}}.\left(c+a\right)\)

        \(\le\sqrt{\frac{3}{2}}.\left(2a+2b+2c\right)=\sqrt{\frac{3}{2}}.6=\sqrt{216}=6\sqrt{6}\)Vì a+b+c=6

Dấu = xảy ra khi a=b=c=2

Vây ......

\(-3x^2+15=0\Leftrightarrow-3x^2=-15\Leftrightarrow x^2=5\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

     \(5x^2-20=0\)

\(\Leftrightarrow5x^2=20\)

\(\Leftrightarrow x^2=4\)

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

dyttt cuuu may x=20320302

\(2x^3-7x^2+4x+1=0\)

\(\Leftrightarrow2x^2\left(x-1\right)-5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^2-5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x^2-5x-1=0\end{cases}}\) Đến đây tự làm tiếp nha

tìm x à viet trung nguyen

\(7x^2-5x=0\)

\(\Leftrightarrow x\left(7x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{7}\end{cases}}\)

\(a)\) Ta có : 

\(M=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\)

Thay \(a+b=1\) vào \(M=\left(a+b\right)\left(a^2+b^2-ab\right)\) ta được : 

\(M=\left(a+b\right)\left(a^2+b^2-ab\right)=1\left(a^2+b^2-ab\right)=a^2+b^2-ab\)

Lại có : 

\(a^2\ge0\)

\(b^2\ge0\)

\(\Rightarrow\)\(a^2+b^2\ge0\)

\(\Rightarrow\)\(a^2+b^2-ab\ge-ab\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a^2=0\\b^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=0\\b=0\end{cases}}}\)

Vậy \(M_{min}=-ab\) khi \(a=b=0\)

Sai thì thôi nhé, mk mới lớp 7 

dytt me dễ vãi lone

\(a^3+\frac{1}{8}+\frac{1}{8}\ge3\sqrt[3]{\frac{a^3.1}{8.8}}=\frac{3}{4}a.\)

\(b^3+\frac{1}{8}+\frac{1}{8}\ge\frac{3}{4}b\)

\(M+\frac{4}{8}\ge\frac{3}{4}\left(a+b\right)=\frac{3}{4}\Leftrightarrow M\ge\frac{3}{4}-\frac{4}{8}=?\) tự tính dcmmm

b.

\(a^3+1+1\ge3\sqrt[3]{a^3}=3a\)

\(b^3+1+1\ge3b\)

\(a^3+b^3+4\ge3\left(A+b\right)\)

cái dmcmmm a^3+b^3=2 suy ra

\(6\ge3\left(a+b\right)\)

\(2\ge a+b\)

dytt cụ m tự kết luận