chịu rồi bạn ạ

\(Taco:\)

\(A=2\left(3x+1\right)\left(x-1\right)-3\left(2x-3\right)\left(x-4\right)\)

\(A=\left(6x+2\right)\left(x-1\right)-\left(6x-9\right)\left(x-4\right)\)

\(A=\left(6x^2-4x-2\right)-\left(6x^2-24x-9x-36\right)\)

\(A=6x^2-4x-2-6x^2+33x+36=29x+34\)

\(b,x=2\Rightarrow A=58+34=92\)

\(A=-20\Leftrightarrow29x=-20-34=-54\Leftrightarrow x=\frac{-54}{29}\)

\(x^2\ge0.\Rightarrow A+x^2=x\left(x+29\right)+34\ge-176,25\)

Dấu "=" xảy ra khi: x(x+29) đạtGTNN

<=> x=-14,5

Ta có: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

\(\Leftrightarrow x^2+2xy+y^2=z^2\)

\(\Leftrightarrow x^2+y^2-z^2=-2xy\)

Chứng minh tương tự ta có:

\(x^2+z^2-y^2=-2xz\)

\(y^2+z^2-x^2=-2yz\)

\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)

\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)

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

\(=-\frac{3}{2}\)

Vậy giá trị biểu thức là \(-\frac{3}{2}\)

\(5x^3-5x=5x\left(x^2-1\right)\)

\(3x^2+5x-3xy-5x=x\left(3x+5\right)-x\left(3y+5\right)=x\left(3x-3y\right)=3x\left(x-y\right)\)

\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)\)

\(=\frac{1}{5}x^2y^2\left(15xy-5+3x\right)\)

\(=\frac{1}{5}\left(x.y\right)^2.\left(15xy-5+3x\right)\)

\(=\frac{1}{5}\left(15x^3y^3-5x^2y^2+3x^3y^2\right)\)

\(a,4x^2-y^2-1-4x\)

\(\Rightarrow\left(4x^2+4x+1\right)-y^2\)

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

\(b,6x^2-7x-20=6x^2-15x+8x-20\)

\(=\left(6x^2-15x\right)+\left(8x-20\right)\)

\(=3x\left(2x-5\right)+4\left(2x-5\right)\)

\(=\left(3x+4\right)\left(2x-5\right)\)

a) Hình thành nhiều đới, niều kiểu khí hậu khác nhau

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