Câu 1 : Tìm x :

1. \(A=x^2+4x-2\)

\(A=x^2+2.x.2+2^2-2^2-2\)

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

\(A=\left(x+2\right)^2-6\)

\(\left(x+2\right)^2-6\ge-6\)

MIn A= -6 khi \(\left(x+2\right)^2=0\)

=> \(x+2=0hayx=-2\)

Vậy x=2

những câu tiếp theo làm tg tự như thế nhé

Câu 1:

a) Ta có: \(A=x^2+4x-2\)

\(=x^2+4x+4-6\)

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

Ta có: \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2-6\ge-6\forall x\)

Dấu '=' xảy ra khi

\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy: x=-2

b) Ta có: \(B=2x^2-4x+3\)

\(=2\left(x^2-2x+\frac{3}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot1+1+\frac{1}{2}\right)\)

\(=2\left[\left(x^2-2x\cdot1+1\right)+\frac{1}{2}\right]\)

\(=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)

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

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi

\(2\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy: x=1

c) Ta có: \(C=x^2+y^2-4x+2y+5\)

\(=x^2-4x+4+y^2+2y+1\)

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

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\left(y+1\right)^2\ge0\forall y\)

Do đó: \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

Vậy: x=2 và y=-1

Câu 2:

a) Ta có: \(A=-x^2+6x+5\)

\(=-\left(x^2-6x-5\right)\)

\(=-\left(x^2-6x+9-14\right)\)

\(=-\left[\left(x^2-6x+9\right)-14\right]\)

\(=-\left[\left(x-3\right)^2-14\right]\)

\(=-\left(x-3\right)^2+14\)

Ta có: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x-3\right)^2+14\le14\forall x\)

Dấu '=' xảy ra khi

\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy: GTLN của đa thức \(A=-x^2+6x+5\) là 14 khi x=3

b) Ta có: \(B=-4x^2-9y^2-4x+6y+3\)

\(=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2-5\right]\)

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

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\)(1)

Ta có: \(\left(3y-1\right)^2\ge0\forall y\)

\(\Rightarrow-\left(3y-1\right)^2\le0\forall y\)(2)

Từ (1) và (2) suy ra

\(-\left(2x+1\right)^2-\left(3y-1\right)^2\le0\forall x,y\)

\(\Rightarrow-\left(2x+1\right)^2-\left(3y-1\right)^2+5\le5\forall x,y\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}-\left(2x+1\right)^2=0\\-\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

Vậy: GTLN của đa thức \(B=-4x^2-9y^2-4x+6y+3\) là 5 khi và chỉ khi \(x=\frac{-1}{2}\) và \(y=\frac{1}{3}\)

Câu 3:

a) Ta có: \(x^2+y^2-2x+4y+5=0\)

\(\Rightarrow x^2-2x+1+y^2+4y+4=0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy: x=1 và y=-2

b) Ta có: \(5x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow x^2+4x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow\left(4x^2+12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

\(\Rightarrow\left(2x+3y\right)^2+\left(x-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)

Vậy: x=3 và y=-2

1,

\(x^2+y^2+y^2=14\)

\(\Rightarrow\left(x+y+z\right)^2-2xy-2yz-2zx=14\)

\(\Rightarrow-2\left(xy+yz+zx\right)=14\)

\(\Rightarrow xy+yz+zx=-7\)

\(\Rightarrow\left(xy+yz+zx\right)^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2+2xyz\left(x+y+z\right)=49\)

\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2=49\)

Ta có: \(x^4+y^4+z^4\)

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

\(=14^2-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(=14^2-2.49\)

\(=196-98\)

\(=98\)

21. Phân tích A thành \(A=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(a^2+b^2+c^2+ab+bc+ac\right)\). Từ đó dễ dàng chứng minh.

23. \(9y\left(y-x\right)=4x^2\Leftrightarrow9y^2-9xy=4x^2\Leftrightarrow4x^2+9xy-9y^2=0\)

Chia cả hai vế của đẳng thức trên với \(y^2>0\)được : 

\(4\left(\frac{x}{y}\right)^2+\frac{9x}{y}-9=0\). Đặt \(t=\frac{x}{y},t>0\)(Vì x,y dương)

\(\Rightarrow4^2+9t-9=0\Leftrightarrow\orbr{\begin{cases}t=\frac{3}{4}\left(\text{nhận}\right)\\t=-3\left(\text{loại}\right)\end{cases}}\)

Vậy \(\frac{x}{y}=\frac{3}{4}\Rightarrow y=\frac{4x}{3}\)thay vào biểu thức được :

\(\frac{x-y}{x+y}=\frac{x-\left(\frac{4x}{3}\right)}{x+\left(\frac{4x}{3}\right)}=-\frac{1}{7}\)

24. Tương tự câu 23 , ta được \(x=y\) hoặc \(y=-3x\)(loại trường hơp này vì mẫu thức phải khác 0)

 Vậy với x = y được \(A=-\frac{1}{2}\)

a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)

\(=\left(-6x-18\right)\left(8x^2-18\right)\)

\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)

\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)

b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)

\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)

\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)

c) Ta có: \(-4x^2+12xy-9y^2+25\)

\(=-\left(4x^2-12xy+9y^2-25\right)\)

\(=-\left[\left(2x-3y\right)^2-25\right]\)

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

d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)

\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)

\(=\left(x-y\right)^2-\left(2m-n\right)^2\)

\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)

a) (4x²-3x-18)²-(4x²+3x)²

=(4x²-3x-18-4x²-3x)(4x²-3x-18+4x²+3x)

=(-6x-18)(8x²-18)

=-48x³+108x-144x²+324

Bài 4: 

a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)

\(\Leftrightarrow3x-40=2\)

=>3x=42

hay x=14

b: \(\Leftrightarrow x^3+8-x^3-2x=0\)

=>-2x+8=0

=>-2x=-8

hay x=4

c: \(x\left(x-2\right)+\left(x-2\right)=0\)

=>(x-2)(x+1)=0

=>x=2 hoặc x=-1

d: \(5x\left(x-3\right)-x+3=0\)

=>5x(x-3)-(x-3)=0

=>(x-3)(5x-1)=0

=>x=3 hoặc x=1/5

e: \(3x\left(x-5\right)-\left(x-1\right)\left(3x+2\right)=30\)

\(\Leftrightarrow3x^2-15x-3x^2-2x+3x+2=30\)

=>-14x=28

hay x=-2

f: \(\Leftrightarrow\left(x+2\right)\left(x+30-x-5\right)=0\)

=>x+2=0

hay x=-2