a)=1-4a
b) = 2x - 4y
c) = 2x - 2 (nếu x>5)
=2x(nếu x<5)
 

-1.       2x.        2x

a) A= 1 - 4a
b) B=2x-4y
c) C= 2x-2 (nếu x>5)
  = 2x (nếu x< 5)

Lời giải:

a)

\(\sqrt{1-4a+4a^2}-2a=\sqrt{1-2.2a+(2a)^2}-2a\)

\(=\sqrt{(2a-1)^2}-2a=|2a-1|-2a=(2a-1)-2a=-1\)

(do $a\geq \frac{1}{2}$ nên $|2a-1|=2a-1$)

b)

\(x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{(x-2y)^2}=x-2y-|x-2y|\)

\(=x-2y-(2y-x)=2(x-2y)\)

(do $x< 2y$ nên $|x-2y|=-(x-2y)=2y-x$)

c)

\(x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{(x^2)^2-2.4.x^2+4^2}\)

\(=x^2+\sqrt{(x^2-4)^2}=x^2+|x^2-4|=x^2+(4-x^2)=4\)

(do $x^2< 4$ nên $|x^2-4|=4-x^2$)

bổ sung: ý a) điều kiện x<2

\(a,\sqrt{1-4a+4a^2}-2a\)

\(=\sqrt{\left(1-2a\right)^2}-2a\)

\(=1-2a-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

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

\(=x-2y-\left(x-2y\right)\)

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

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\sqrt{\left(x^2-4\right)^2}\)

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

\(=2x^2-4\)

Các câu còn lại tương tự nha

\(a,\sqrt{1-4a+4a^2}-2a\)

\(=\sqrt{\left(1-2a\right)^2}-2a\)

\(=\left(1-2a\right)-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

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

\(=x-2y-\left(x-2y\right)\)

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

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

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

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

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

\(=2x^2-4\)

\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)

\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)

\(=2x-1-\frac{x-5}{x-5}\)

\(=2x-1-1\)

\(=2x-2\)

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

Bài 1

a) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x-1\right)\left(x+1\right)\)

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

\(=3x^3+6x-3x^3+3x=9x\)

b) \(\left(a+b+c\right)^2+\left(a+b-c\right)^2+\left(2a-b\right)^2\)

\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2bc-2ca+4a^2-4ab+b^2\)

\(=6a^2+3b^2+2c^2+4ab-4ab=6a^2+3b^2+2c^2\)

Bài 2 

a) \(x^2-20x+101=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

Dấu = xảy ra \(< =>\left(x-10\right)^2=0< =>x-10=0< =>x=10\)

b) \(4a^2+4a+2=4\left(a^2+a+\frac{1}{4}\right)+1=4\left(a+\frac{1}{2}\right)^2+1\ge1\)

Dấu = xảy ra \(< =>4\left(a+\frac{1}{2}\right)^2=0< =>a+\frac{1}{2}=0< =>a=-\frac{1}{2}\)

c) \(x^2-4xy+5y^2+10x-22y+28=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+y^2-2y+1+27\)

\(=\left(x-2y\right)^2+2.5.\left(x-2y\right)+25+\left(y-1\right)^2+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu = xảy ra \(< =>\hept{\begin{cases}y-1=0\\x-2y+5=0\end{cases}< =>\hept{\begin{cases}y=1\\x=-3\end{cases}}}\)

Bài 3 

a) \(4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Dấu = xảy ra \(< =>\left(x-2\right)^2=0< =>x-2=0< =>x=2\)

b) \(x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

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

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé. Viết thế này khó đọc quá.

\(x+2y-\sqrt{x^2-4xy+4y^2}\)(sửa đề)

\(=x+2y-\sqrt{\left(x-2y\right)^2}\)

\(=x+2y-\left|x-2y\right|\)

\(=x+2y-\left(x-2y\right)\left(vì.x\ge2y\right)\)

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

\(=4y\)

\(x+2y-\sqrt{x^2-4xy+4y^2}^2\)

\(=x+2y-\sqrt{\left(x-2y\right)^2}^2\)

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

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