Bài làm:

Ta có: \(\frac{x^3}{8}+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)

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

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

a, \(A=x\left(x+y\right)-x\left(y-x\right)=x^2+xy-xy+x^2=2x^2\)

Thay x vào ta có : \(2\left(-3\right)^2=2.9=18\)

y bị lược bỏ rồi mà bạn hay chỗ x^2 + xy - xy + x^2 thay vào à ? lạ !?!

b, \(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)=8x^2+4xy+4xy+2y^2-y^2-2xy\)

\(=8x^2+6xy+y^2\)

Thay x = 1/2 ; y = -3/4 ta có : Tự thay nhé -> P/s 

a)

\(A=x.\left(x+y\right)-x.\left(y-x\right)\)

\(A=x^2+x.y-x.y+x^2\)

\(A=2.x^2\)

Thay x= -3 vào biểu thức A ta được ;

\(A=2.\left(-3\right)^2=2.9=18\)

b) \(B=4.x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)\)

\(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(2x+y\right)\)

\(B=\left(2x+y\right).\left(4x+2y-y\right)\)

\(B=\left(2x+y\right).\left(4x+y\right)\)

\(B=8x^2+2xy+4xy+y^2\)

\(B=8x^2+6xy+y^2\)

Thay \(x=\frac{1}{2}\) và \(y=\frac{-3}{4}\) vào biểu thức B ta được :

\(B=8.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}.\left(\frac{-3}{4}\right)+\left(\frac{-3}{4}\right)^2\)

\(B=2+\left(\frac{-9}{4}\right)+\frac{9}{16}=\frac{5}{16}\)

Bài 2 :

\(A=4\left(x-6\right)-5x\left(x+1\right)+8\left(x^2-x-2\right)\)

\(A=4x-24-5x^2-5x+8x^2-8x-16\)

\(A=-9x-40+3x^2\)

Thay x=-1 vào biểu thức A ta được :

\(A=-9.\left(-1\right)-40+3.\left(-1\right)^2\)

\(A=9-40+3=-28\)

Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^

\(\)

\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=0\)

<=>\(\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}=0\)

<=>\(\frac{y^2}{xy\left(2x-y\right)}-\frac{4x^2}{xy\left(2x-y\right)}=0\)

 =>y²-(2x)²=0

<=>(y-2x)(y+2x)=0

<=>y-2x=0 hoặc y+2x=0

M chỉ làm đc đến đó thôi!!!!!

Giải BPT?

Ta có: \(x^2-4x+1< 0\)

\(\Leftrightarrow\left(x^2-4x+4\right)-3< 0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{3}\right)^2< 0\)

\(\Leftrightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)< 0\)

Nhận thấy \(x-2-\sqrt{3}< x-2+\sqrt{3}\)

=> \(\hept{\begin{cases}x-2-\sqrt{3}< 0\\x-2+\sqrt{3}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2+\sqrt{3}\\x>2-\sqrt{3}\end{cases}}\)

Vậy \(2-\sqrt{3}< x< 2+\sqrt{3}\)

Mình làm tắt thôi nhé

\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x^2-x+1\right)}=\frac{\left(x-1\right)^2}{x^2-x+1}\left(x\ne-1\right)\)

Dễ thấy \(A\ge0\)

\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1}{x^4-x^3+x^2+2x^2-2x^2+2x+x^2-x+1}\)

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

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

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

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

Ta có : \(\frac{\left(x-1\right)^2}{x^2-x+1}=\frac{\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\)

=> Đpcm

Ai giúp em đi ạ

Em hứa sẽ vote nhiệt tình luôn huhuuu

a, (x² - 3)(2 + 4x)

= 2x² + 4x³ - 6 - 12x

b, 6(3x + 5)(x - 4)

= (18x + 30)(x - 4)

= 18x² - 72 + 30x - 120

= 18x² + 30x - 192

\(x^4+2x^3+3x^2+2x=y^2-y\)

\(\Leftrightarrow x^4+x^2+1+2x^3+2x^2+2x=y^2-y+1\)

\(\Leftrightarrow\left(x^2+x+1\right)^2=\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\Leftrightarrow\left(x^2+x+1-y+\frac{1}{2}\right)\left(x^2+x+1+y-\frac{1}{2}\right)=\frac{3}{4}\)

\(\Leftrightarrow\left(x^2+x-y+\frac{3}{2}\right)\left(x^2+x+y+\frac{1}{2}\right)=\frac{3}{4}\)

\(\Leftrightarrow\left(2x^2+2x-2y+3\right)\left(2x^2+2x+2y+1\right)=3\)

Đến đây chắc khó.

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

mà \(a^2+b^2=8\)\(\Rightarrow8-2ab=3\)

\(\Rightarrow2ab=5\)\(\Rightarrow ab=\frac{5}{2}\)

Vậy \(ab=\frac{5}{2}\)

b) Ta có: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

mà \(a-b=2\)và \(a+b=4\)

\(\Rightarrow a^2-b^2=2.4=8\)

Vậy \(a^2-b^2=8\)

a) Ta có: \(\hept{\begin{cases}a^2+b^2=8\\\left(a-b\right)^2=3\end{cases}}\Leftrightarrow\hept{\begin{cases}a^2+b^2=8\\a^2-2ab+b^2=3\end{cases}}\)

=> \(a^2+b^2-\left(a^2-2ab+b^2\right)=8-3\)

<=> \(2ab=5\)

=> \(ab=\frac{5}{2}\)

b) Ta có: \(a^2-b^2=\left(a-b\right)\left(a+b\right)=2.4=8\)

lm lộn đề nên hơi chậm xíu^^

1) \(2x^4+5x^2+2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x^2+1=0\\x^2+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=-\frac{1}{2}\\x^2=-2\end{cases}}\) (vô lý)

=> pt vô nghiệm

2) \(2x^4-7x^2-4=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\2x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=-\frac{1}{2}\left(vl\right)\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

3) \(x^4-5x^2+4=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=1\\x^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm1\\x=\pm2\end{cases}}\)

4) \(2x^4-20x^2+18=0\)

\(\Leftrightarrow x^4-10x^2+9=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=9\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm1\\x=\pm3\end{cases}}\)

1. \(2x^4+5x^2+2=0\)

Vì \(2x^4+5x^2+2\ge2\)

=> Pt trên vô nghiệm

2. \(2x^4-7x^2-4=0\)

\(\Leftrightarrow2x^4+x^2-8x^2-4=0\)

\(\Leftrightarrow x^2\left(2x^2+1\right)-4\left(2x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x^2+1\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}2x^2+1=0\left(vo-ly\right)\\x+2=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

Bài làm:

Ta có: \(\hept{\begin{cases}\left(a-b\right)^2=2\\\left(a+b\right)^2=3\end{cases}\Leftrightarrow}\hept{\begin{cases}a^2-2ab+b^2=2\\a^2+2ab+b^2=3\end{cases}}\)

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

<=> \(2\left(a^2+b^2\right)=5\)

=> \(a^2+b^2=\frac{5}{2}\)

Thay vào tính được: \(\frac{5}{2}+2ab=3\Leftrightarrow2ab=\frac{1}{2}\Rightarrow ab=\frac{1}{4}\)

Vậy \(a^2+b^2=\frac{5}{2}\) và \(ab=\frac{1}{4}\)

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

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

Trừ (2) cho (1) ta được: \(\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)=3-2\)

\(\Leftrightarrow4ab=1\)\(\Leftrightarrow ab=\frac{1}{4}\)

\(\Rightarrow a^2+b^2=3-2.\frac{1}{4}=\frac{5}{2}\)

Vậy \(a^2+b^2=\frac{5}{2}\)và \(ab=\frac{1}{4}\)