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1: ĐKXĐ: x<>-3; x<>2
2: \(D=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
3: x^2-9=0
=>x=3(nhận) hoặc x=-3(loại)
Khi x=3 thì \(D=\dfrac{3-4}{3-2}=-1\)
4: Để D=-3/4 thì \(\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
=>4x-16=-3x+6
=>7x=22
=>x=22/7
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=7^3-3.8.7=\)
\(4x^3-13x^2+9x-18\)
\(=4x^3-12x^2-x^2+3x+6x-18\)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
Ta có:
\(x^2+3x-18\)
\(\Leftrightarrow x^2-3x+6x-18\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(6x-18\right)\)
\(\Leftrightarrow x\left(x-3\right)+6\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+6\right)\)
a) \(A=9x^2-12x+4+9x^2+12x+4+18x^2-8=36x^2=36.\left(-\dfrac{1}{3}\right)^2=4\)
b) \(B=\left(x+y-7\right)^2-2\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2=\left(x+y-7-y+6\right)^2=\left(x-1\right)^2=\left(101-1\right)^2=100^2=10000\)
\(A=x^2+10y^2+2x-6xy-10y+25\)
=> \(A=x^2+2x\left(1-3y\right)+\left(1-3y\right)^2-\left(1-3y\right)^2-10y+25\)
=> \(A=\left(x+1-3y\right)^2-1+6y-9y^2-10y+25\)
=> \(A=\left(x+1-3y\right)^2-9y^2-4y+24\)
=> \(A=\left(x+1-3y\right)^2-\left(3y\right)^2-2.3y.\frac{2}{3}-\left(\frac{2}{3}\right)^2+\frac{220}{9}\)
=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\)
Có \(\left(x+1-3y\right)^2\ge0\)với mọi x, y
\(\left(3y+\frac{2}{3}\right)^2\ge0\)với mọi y
=> \(A=\left(x+1-3y\right)^2-\left(3y+\frac{2}{3}\right)^2+\frac{220}{9}\ge\frac{220}{9}\)với mọi x, y
Dấu "=" xảy ra <=> \(\left(x+1-3y\right)^2=0\)<=> \(x+1-3y=0\)
và \(\left(3y+\frac{2}{3}\right)^2=0\)=> \(3y+\frac{2}{3}=0\)
=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)
Bổ xung phần kết luận
KL: Amin = \(\frac{220}{9}\)<=> \(\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{-2}{9}\end{cases}}\)
Oh chịu ròi