Chọn B

bài 1 áp dụng hdt là ra

bài 2 cũng z, nó tòi ra 1 số thì gtnn = cái số đó

bài 3

câu a phá hết ra

câu b nhóm hạng tử

câu a trương tự, trong ngoặc sẽ tạo ra 1 hđt

bài 4 câu a phá hết

câu b hằng đẳng thức

\(A=x^2-6x+10\)

\(=x^2-6x+9+1\)

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

\(\left(x-3\right)^2\ge0\)

\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)

Vậy A > 0 với mọi x.

\(B=x^2-2xy+y^2+1\)

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

\(\left(x-y\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+1\ge1>0\)

Vậy B > 0 với mọi x, y.

\(M=x^2-6x+12\)

\(=x^2-6x+9+3\)

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

\(\left(x-3\right)^2\ge0\)

\(\Rightarrow\left(x-3\right)^2+3\ge3\)

\(MinB=3\Leftrightarrow x=3\)

\(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)

\(x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=7\)

\(2x^2+6x+5-2x^2+4x-2=7\)

\(10x=7+3\)

\(10x=10\)

\(x=1\)

\(x^2+x=0\)

\(x\left(x+1\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)

\(x^3-\frac{1}{4}x=0\)

\(x\left(x^2-\frac{1}{4}\right)=0\)

\(x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)

\(\left(x+10\right)^2-\left(x^2+2x\right)\)

\(=x^2+20x+100-x^2-2x\)

\(=18x+100\)

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

\(=x^2-4+x^3-1-x^3-x^2\)

\(=-5\)

hơi ngán dạng này :((((

a, \(x^2-3x+5=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

b,

\(x^2-\frac{1}{3}x+\frac{5}{4}=x^2-2.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{5}{4}=\left(x-\frac{1}{6}\right)^2+\frac{11}{9}>0\forall x\)

c,

\(x-x^2-3=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}-3=-\left(x-\frac{1}{2}\right)^2-\frac{11}{4}< 0\forall x\)d,

\(x-2x^2-\frac{5}{2}=-2\left(x^2-\frac{1}{2}x+\frac{5}{4}\right)=-2\left(x^2-2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{4}\right)=-2\left[\left(x-\frac{1}{4}\right)^2+\frac{19}{16}\right]=-2\left(x-\frac{1}{4}\right)^2-\frac{19}{8}< 0\forall x\)P/s : ko chắc lém :)))

\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)

a: \(\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\)

\(=\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+35}{x-25}\)

b: \(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\)

\(=\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}=\dfrac{x+5\sqrt{x}-2}{x-9}\)

c: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{x-4}\)

\(=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

d: \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

Bài 1.

a) ( 7x - 3 )² - 5x( 9x + 2 ) - 4x² = 18

<=> 49x² - 42x + 9 - 45x² - 10x - 4x² = 18

<=> -52x + 9 = 18

<=> -52x = 9

<=> x = -9/52 

b) ( x - 7 )² - 9( x + 4 )² = 0

<=> x² - 14x + 49 - 9( x² + 8x + 16 ) = 0

<=> x² - 14x + 49 - 9x² - 72x - 144 = 0

<=> -8x² - 86x - 95 = 0 

<=> -8x² - 10x - 76x - 95 = 0

<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0

<=> ( x + 5/4 )( -8x - 76 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)

c) ( 2x + 1 )² + ( 4x - 1 )( x + 5 ) = 36

<=> 4x² + 4x + 1 + 4x² + 19x - 5 = 36

<=> 8x² + 23x - 4 - 36 = 0

<=> 8x² + 23x - 40 = 0

=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))

Bài 2.

a) x² - 12x + 39 = ( x² - 12x + 36 ) + 3 = ( x - 6 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) 17 - 8x + x² = ( x² - 8x + 16 ) + 1 = ( x - 4 )² + 1 ≥ 1 > 0 ∀ x ( đpcm )

c) -x² + 6x - 11 = -( x² - 6x + 9 ) - 2 = -( x - 3 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

d) -x² + 18x - 83 = -( x² - 18x + 81 ) - 2 = -( x - 9 )² - 2 ≤ -2 < 0 ∀ x ( đpcm )

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

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

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

\(\Leftrightarrow4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

b ) \(\left(2x+3\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

c ) \(x^3-8=\left(x-2\right)^2\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+2x+4-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2=-\dfrac{23}{4}\end{matrix}\right.\) ( Vô lý )

Vậy \(x=2\)

d ) \(x^3+5x^2-4x-20=0\)

\(\Leftrightarrow x^2\left(x+5\right)-4\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x+5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-5\end{matrix}\right.\)

Vậy ...

e ) \(x^3-4x^2+4x=0\)

\(\Leftrightarrow x\left(x^2-4x+4\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy ...

f ) \(x^2-25+2\left(x+5\right)=0\)

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

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

\(\Leftrightarrow\left(x+5\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

g ) Sai đề

h ) \(x^2\left(x-2\right)+7x=14\)

\(\Leftrightarrow x^2\left(x-2\right)+7x-14=0\)

\(\Leftrightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-7\left(VL\right)\\x=2\end{matrix}\right.\)

Vậy \(x=2\)