bài 1, bạn tự làm nhé đặt chia đi bạn

bài 2

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

\(b,=a^2-2a-5a+10=a\left(a-2\right)-5\left(a-2\right)=\left(a-2\right)\left(a-5\right)\)

Gọi thương của phép chia n cho 11 là x 

thì n=11x+4

\(\Leftrightarrow n^2=\left(11x+4\right)^2\)

\(\Leftrightarrow n^2=121x^2+88x+16\)

=> n² :11   <=>\(121x^2+88x+16:11\)

                     \(\Leftrightarrow11\left(11x^2+8x+1\right)+5\)

vậy n² :11 dư 5

b, bạn làm tương tự nhé rồi đặt 13 làm nhân tử chung thì sẽ chia hết cho 13 

Tính
\(\left(0,3x^2y-\frac{16}{3}z\right)^2\) 
b) \(\left[\left(2u+5\right)-3z\right]^2\)

Bình phương a và b lên để so sánh

b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)

c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)

a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)

\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)

\(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}\)

\(=\frac{a+c}{a+b}+\frac{2c}{a+b}+\frac{a+b}{a+c}+\frac{2b}{a+c}+\frac{2a}{b+c}\)

\(=2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)\)

Áp dụng BĐT Cauchy - Schwar:

\(\frac{a+c}{a+b}+\frac{a+b}{a+c}\ge2\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{\left(a+b\right)\left(a+c\right)}}=2\)(1)

Áp dụng BĐT Nesbit:

\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\ge\frac{3}{2}\)

\(\Leftrightarrow2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)\ge3\)(2)

Từ (1) và (2) suy ra \(2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)\ge5\)

hay \(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}\ge\left(đpcm\right)\)

Ta có: \(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}-5\ge0\)

\(\Leftrightarrow\frac{a+3c}{a+b}-2+\frac{a+3b}{a+c}-2+\frac{2a}{b+c}-1\ge0\)

Giải bất phương trình

Cuối cùng ta được: \(\left(c-a\right)^2\left(\frac{1}{\left(a+b\right)\left(b+c\right)}\right)+2\left(b-c\right)^2\left(\frac{1}{\left(a+c\right)\left(a+b\right)}\right)+\left(a-b\right)^2\) \(\left(\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\ge0\)

BĐT đúng <=> a = b = c

\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)

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

\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)

\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}y^4+\left(y^4\right)^2=\left(\frac{1}{3}-y^4\right)^2\)

a,(x-3)^2

b,(1/4x+2b^2)^2

c,(5+x)^2

d,(1/3-y^4)^2

a) x(2x+1)-x²(x+2)+(x³-x+3)

=2x²+ x- x³-2x²+ x³-x+3

=3

b) x(3x² -x +5)- ( 2x³ +3x- 16)-x(x²- x+2)

=3x³ - x² + 5x- 2x³ -3x +16- x³+x²-2x

=16

a)\(\left(x+1\right)\left(x+3\right)-x\left(x-2\right)=7\)

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

\(x^2+3x+x+3-x^2+2x=7\)

\(6x+3=7\)

\(6x=4\)

\(x=\frac{4}{6}=\frac{2}{3}\)

Vậy  \(x=\frac{2}{3}\)

b) \(2x\left(3x+5\right)-x\left(6x-1\right)=33\)

\(6x^2+10x-6x^2-x=33\)

\(9x=33\)

\(x=\frac{33}{9}\)

Vậy \(x=\frac{33}{9}\)