Chỗ cuối mình viết nhầm nha là \(64x^4\)

Võ Thị Thảo Minh             

em hãy sử dụng đẳng thức này để rút gọn :

 a² - b² = (a - b)(a + b)

a) \(x^2-20x+101\)

\(=-\left(x^2+20x-101\right)\)

\(=-\left[\left(x^2+2x.10-10^2\right)+1\right]\)

\(=\left[\left(x-10\right)^2+1\right]\)

\(=-\left(x-10\right)^2-1\)

Nhận xét : \(-\left(x-10\right)^2\le0\)với mọi x

\(\Leftrightarrow-\left(x-10\right)^2-1\le-1\) với mọi x

Vậy GTLN của biểu thức là -1 đạt được khi :

(x-10)² = 0

=> (x-10) =0

=> x = 0 + 10

=> x = 10

~Chắc vậy~

b/ \(4x^2+4x+2\)

= \(\left[\left(2x\right)^2+2.2x.1+1^2\right]+1\)

= \(\left(2x+1\right)^2+1\) \(\ge1\forall x\in R\)

Dấu '' = '' xảy ra <=> \(\left(2x+1\right)^2=0\) => \(x=\dfrac{-1}{2}\)

Vậy Max_B = 1 <=> \(x=\dfrac{-1}{2}\)

a: Ta có: \(10x^4-27x^3y-110x^2y^2-27xy^3+10y^4\)

\(=10x^4+20x^2y^2+10y^4-27xy\left(x^2+y^2\right)-130x^2y^2\)

\(=10\left(x^2+y^2\right)^2-27xy\left(x^2+y^2\right)-130x^2y^2\)

\(=10\left(x^2+y^2\right)^2-52xy\left(x^2+y^2\right)+25xy\left(x^2+y^2\right)-130x^2y^2\)

\(=2\left(x^2+y^2\right)\left(5x^2+5y^2-26xy\right)+5xy\left(5x^2+5y^2-26xy\right)\)

\(=\left(5x^2-26xy+5y^2\right)\left(2x^2+5xy+2y^2\right)\)

\(=\left(5x^2-25xy-xy+5y^2\right)\left(2x^2+4xy+xy+2y^2\right)\)

\(=\left\lbrack5x\left(x-5y\right)-y\left(x-5y\right)\right\rbrack\left\lbrack2x\left(x+2y\right)+y\left(x+2y\right)\right\rbrack\)

=(5x-y)(x-5y)(2x+y)(x+2y)

b: \(x^5-4x^4+3x^3+3x^2-4x+1\)

\(=x^5+x^4-5x^4-5x^3+8x^3+8x^2-5x^2-5x+x+1\)

\(=\left(x+1\right)\left(x^4-5x^3+8x^2-5x+1\right)\)

\(=\left(x+1\right)\left(x^4-x^3-4x^3+4x^2+4x^2-4x-x+1\right)\)

\(=\left(x+1\right)\left(x-1\right)\left(x^3-4x^2+4x-1\right)\)

\(=\left(x+1\right)\left(x-1\right)\left\lbrack\left(x^3-x^2\right)-3x^2+3x+x-1\right\rbrack\)

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

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\left(ĐK:x\ne0;x\ne-4\right)\)

\(=\frac{6}{x\left(x+4\right)}+\frac{3}{2\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(4+x\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)

b) \(\frac{4xy-5}{140x^3y}-\frac{6y^2-5}{10x^3y}\left(ĐK:x,y\ne0\right)\)

\(=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{2y\left(2x-3y\right)}{10x^3y}=\frac{2x-3y}{5x^3}\)

- cảm ơn bạn nhiều nha !

\(4x^4-10x^3+8x^2-5x-1=0\)

\(\left(x^4-x^3+2x^2\right)-\left(4x^3-4x^2+8x\right)+\left(2x^2-2x+4\right)=0\)

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

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

\(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]\left(x^2-4x+2\right)=0\)

Vì \(\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]>0\)\(\Rightarrow x^2-4x+2=0\)

\(\Rightarrow\left(x-2\right)^2=2\)\(\Rightarrow x-2=\pm\sqrt{2}\)

\(\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+2\\x=2-\sqrt{2}\end{cases}}\)