a: Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)

Bài 7:

a: a>b

nên -2a<-2b

=>-2a-6<-2b-6<-2b

b: a>b

nên 3a>3b

=>3a-9>3b-9

=>3(a-3)>3(b-3)

1 If we continue polluting the air, the environment will be affected seriously

2 If we don't stop damaging birds' homes, there will be fewer birds left

3 If we continue polluting the water, there will be more disease in people and animals

4 If we keep polluting the land, the soil will not be safe enough for us to grow food in

5 If we use public transportation when we need, there will be less air pollution

12 Daniel signs up for a clean-up campaign so that he can join in it

13 People in my neighborhood clean the park up so that their children will have a nice place to play

14 Almost all the students in my school plant trees so that they can have much fresher air

Cái chỗ này mình xin lỗi bạn nhiều nha, mình bị sai chỗ này rồi

Ta có: \(\left(a-b\right)^2>=0\forall a,b\)

=>\(a^2+b^2-2ab>=0\forall a,b\)

=>\(a^2+b^2>=2ab\forall a,b\)

Dấu "=" xảy ra khi a=b

\(2\cdot\left(\dfrac{a}{2}\cdot\dfrac{b}{2}\right)< =\left(\dfrac{a}{2}\right)^2+\left(\dfrac{b}{2}\right)^2=\dfrac{a^2+b^2}{4}\)

=>\(2\cdot\left(2\cdot\dfrac{a}{2}\cdot\dfrac{b}{2}\right)< =\dfrac{2\left(a^2+b^2\right)}{4}=\dfrac{a^2+b^2}{2}\)

Bạn bỏ giúp mình dấu căn nha

+)x=0 khong phai la nghiem cua phuong trinh

+)chia ca 2 ve cho \(x^2\ne\) 0 ta co:

\(x^2-5x+8-\frac{5}{x}+\frac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+8=0\)         (1)

Dat \(x+\frac{1}{x}=a\)   \(\left(\left|a\right|\ge2\right)\)

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

(1)\(\Leftrightarrow\)\(\left(a^2-2\right)-5a+8=0\)

den day ban tu giai tiep nhe

Bài 1: 

a: ĐKXĐ: \(x\ge\dfrac{3}{2}\)

Bài 4:

\(a,A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ P=A:B=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\\ b,P\sqrt{x}=m-\sqrt{x}+x\\ \Leftrightarrow x-1=m-\sqrt{x}+x\\ \Leftrightarrow m=\sqrt{x}-1\)

a) 3x-9xy=3.x-3x.3y

Bài 2: 

a: \(2x^4-8x^2=0\)

\(\Leftrightarrow2x^2\left(x^2-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b: Ta có: \(\dfrac{2}{5}x\left(x+10\right)-x-10=0\)

\(\Leftrightarrow\left(x+10\right)\left(\dfrac{2}{5}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=\dfrac{5}{2}\end{matrix}\right.\)

bài 1:

a) 3x-9xy

= 3x(1-3y)

b)\(18x^2y^2-9x^3y+36xy^2\) 

= \(9xy\left(2xy-x^2+4y\right)\)

c) \(5y\left(y-x\right)+15x\left(x-y\right)\)

= \(5y\left(y-x\right)-15x\left(y-x\right)\)

= \(\left(y-x\right)\cdot\left(5y-15x\right)\)

= \(\left(y-x\right)\cdot5\left(y-3x\right)\)