F=x²+2xy+y²-x-y-12 

= (x + y)^2 - (x + y) - 12 

= (x + y)(x + y - 1) - 12

đặt x + y = t

F = t(t - 1) - 12

= t^2 - t - 12

=  (t - 4)(t + 3)

G=(x²-3x-1)²-12(x²-3x-1)+27

đăth x^2 - 3x - 1 = t

G = t^2 - 12t + 27

= (t - 3)(t - 9)

có t = x^2 - 3x - 1

thay vào 

Câu F ( kiểm tra lại đề )

 Câu G . Đặt x^2 -3x-1=t

 t^2 -12t+27 ( thực hiện pp tách)

a)Ta có:\(x-y=2\Rightarrow\left(x-y\right)^2=4\Rightarrow\left(x^2+y^2\right)-2xy=4\Rightarrow4-2xy=4\Rightarrow2xy=0\Rightarrow xy=0\)

Khi đó ta có:\(x^5y=xy^5=xy\left(x^4-y^4\right)=0\)

khó ha

\(bđt< =>\frac{a+b}{ab}\ge\frac{4}{a+b}\)

\(< =>a^2+2ab+b^2\ge4ab\)

\(< =>a^2+b^2\ge2ab\)

\(< =>\left(a-b\right)^2\ge0\)*đúng*

Vậy ta có điều phải chứng minh

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Ta có : \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}\ge0\)

\(\frac{b}{ab}+\frac{a}{ab}-\frac{4}{a+b}\ge0\)

\(\frac{a+b}{ab}-\frac{4}{a+b}\ge0\)

\(\frac{\left(a+b\right)^2}{ab\left(a+b\right)}-\frac{4ab}{ab\left(a+b\right)}\ge0\)

\(a^2+2ab+b^2-4ab\ge0\Leftrightarrow\left(a-b\right)^2\ge0\)

Đăngr thức xảy ra <=> a = b 

,[x-y+z]^2+[z-y]^2+2.[x-y+z][y-z] (x - y + z)² + (z - y)² + 2(x - y + z)(y - z)
= (x - y + z)² + 2(x - y + z)(y - z) + (y - z)²
= (x - y + z + y - z)²
= x²

Ta có:

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

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

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

\(=x^2\)

Học tốt nhé 

Ta có: \(y'=a\)\(cosx-b\)\(sinx+1\)

y đồng biến trên R \(\Leftrightarrow y'\ge0,\forall x\in R\)

                         \(\Leftrightarrow acosx-bsinx+1\ge0,\forall x\in R\)(*)

Theo bất đẳng thức Schwartz thì:

\(|acosx-bsinx|\le\sqrt{a^2+b^2},\forall x\)

\(\Leftrightarrow-\sqrt{a^2+b^2}\le acos-bsinx\le\sqrt{a^2+b^2},\forall x\)

\(\Leftrightarrow1-\sqrt{a^2+b^2}\le acos-bsinx+1\le1+\sqrt{a^2+b^2},\forall x\)

Do đó (*) \(\Leftrightarrow1-\sqrt{a^2+b^2}\ge0\)

\(\Leftrightarrow\sqrt{a^2+b^2}\le1\)

\(\Leftrightarrow a^2+b^2\le1\)

Ta có : \(h_1=13,6h_2\)=> \(\frac{h_1}{h_2}=\frac{13,6}{1}\)hay \(\frac{h_1}{13,6}=\frac{h_2}{1}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : \(\frac{h_1}{13,6}=\frac{h_2}{1}=\frac{h_1+h_2}{13,6+1}=\frac{0,44}{14,6}=\frac{11}{365}\)

=> \(h_1=\frac{11}{365}\cdot13,6=\frac{748}{1825}\)

\(h_2=\frac{11}{365}\cdot1=\frac{11}{365}\)

P/S : Đề như thế nào?Số dữ quá 

Ta có :

\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^2=\left(-a\right)^2\)

\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)

Tương tự  \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

Mặt khác \(\left(b+c\right)^2=\left(-a\right)^2\Rightarrow b^3+3bc\left(b+c\right)+c^3=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow P=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ba}=\frac{a^3+b^3+c^3}{2abc}\)

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

Vậy P = 3/2 

P=3/2

Ta có:\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc=3abc\)(Vì a+b+c=0)

Khi đó ta có:\(N=\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ba}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)

Vậy N=3

\(N=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}\)

\(< =>N-1=\frac{a^3+b^3+c^3-abc}{abc}=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{abc}\)

Do \(a+b+c=0\)\(< =>N-1=\frac{0.\left(a^2+b^2+c^2-ab-bc-ca\right)}{abc}=0\)

\(< =>N=1\)