Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9\left(a+b+c\right)}{\left(a+b+c\right)}=9\)

Dấu " = " khi a = b = c

Bài 1:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)

\(\Leftrightarrow a^2y^2+b^2x^2-2abxy=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\)

\(\Leftrightarrow ay=bx\)

\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\)

\(\Rightarrowđpcm\)

Bài 2:

Ta có: \(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)

\(=\left(5a-3b\right)^2-64c^2\)

\(=25a^2-30ab+9b^2-64c^2\)

\(=25a^2-30ab+9b^2-16a^2+16b^2\left(a^2-b^2=4c^2\right)\)

\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2=VP\)

\(\Rightarrowđpcm\)

Mình sẽ thắp một que diêm

L-I-K-E MÌNH NHA Bùi Cao Minh

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

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

Sửa đề: \(10x^2+20xy+10y^2-90\)

\(=10\left(x^2+2xy+y^2-9\right)\)

\(=10\left[\left(x+y\right)^2-9\right]\)

\(=10\left(x+y+3\right)\left(x+y-3\right)\)

Sửa đề: \(10x^2+20xy+10y^2-90\)

\(=10\left(x^2+2xy+y^2-9\right)\)

\(=10\left[\left(x+y\right)^2-9\right]\)

\(=10\left(x+y-3\right)\left(x+y+3\right)\)

\(VT=\left(m-a\right)^2+\left(2m-b\right)^2+\left(3m-c\right)^2\)

\(=m^2-2am+a^2+4m^2-4bm+9m^2-6mc+c^2\)

\(=14m^2-2m\left(a+2b+3c\right)+a^2+b^2+c^2\)

\(=14m^2-14m^2+a^2+b^2+c^2\) ( do \(a+2b+3c=7m\) )

\(=a^2+b^2+c^2=VP\)

\(\Rightarrowđpcm\)

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

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

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

b, \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)

\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(36-96x+64x^2\right)+17\)

\(=5x^2-20-18+48x-32x^2+17\)

\(=-27x^2+48x-21\)

Sửa:\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=15\)

Vậy x = 15