a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

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

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

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

d.

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

e.

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

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

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

b: 1/2x-4=0

=>1/2x=4

hay x=8

a: x+7=0

=>x=-7

e: 4x²-81=0

=>(2x-9)(2x+9)=0

=>x=9/2 hoặc x=-9/2

g: x²-9x=0

=>x(x-9)=0

=>x=0 hoặc x=9

a)\(x+7=0=>x=-7\)

b)\(\dfrac{1}{2}x-4=0=>\dfrac{1}{2}x=4=>x=8\)

c)\(-8x+20=0=>-8x=-20=>x=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100=>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

e)\(4x^2-81=0=>4x^2=81=>x^2=\dfrac{81}{4}=>\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

f)\(x^2-7=0=>x^2=7=>x=\sqrt{7}\)

g)\(x^2-9x=0=>x\left(x-9\right)=0=>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

a: x+7=0

nên x=-7

b: x-4=0

nên x=4

c: -8x+20=0

=>-8x=-20

hay x=5/2

d: x²-100=0

=>(x-10)(x+10)=0

=>x=10 hoặc x=-10

a) x +7 =0

=>x = -7

b) x - 4 =0=>x = 4

c) -8x + 20 = 0 =>-8x =-20 =>\(x=-\dfrac{20}{-8}=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

e)\(4x^2-81=0=>4x^2=81=>x^2=\dfrac{81}{4}=>\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

f)\(x^2-7=0=>x^2=7=>x=\sqrt{7}\)

g)\(x^2-9x=0=>x\left(x-9\right)=0=>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

H)\(x^3+3x=0=>x\left(x^2 +3\right)=0=>\left[{}\begin{matrix}x=0\\x^2=-3\left(vl\right)\end{matrix}\right.\)

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

*Tìm giá trị nhỏ nhất

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

Ta có: \(A=x^2-4x+1\)

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

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

\(\Rightarrow\left(x-2\right)^2-5\ge-5\forall x\)

Dấu '=' xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy: Giá trị nhỏ nhất của biểu thức \(A=x^2-4x+1\) là -5 khi x=2

b) \(B=4x^2+4x+11\)

Ta có: \(B=4x^2+4x+11\)

\(=\left(2x\right)^2+2\cdot2x\cdot1+1+10=\left(2x+1\right)^2+10\)

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

\(\Rightarrow\left(2x+1\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi \(\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)

Vậy: Giá trị nhỏ nhất của biểu thức \(B=4x^2+4x+11\) là 10 khi \(x=\frac{-1}{2}\)

*Tìm giá trị lớn nhất

e) \(E=5-8x-x^2\)

Ta có: \(E=5-8x-x^2\)

\(=-\left(-5+8x+x^2\right)=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)=-\left(x+4\right)^2+21\)

Ta có: \(\left(x+4\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x+4\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi \(\left(x+4\right)^2=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

Vậy: Giá trị lớn nhất của biểu thức \(E=5-8x-x^2\) là 21 khi x=-4

f) \(F=4x-x^2+1\)

Ta có: \(F=4x-x^2+1\)

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

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

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

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

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

\(\Rightarrow-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy: Giá trị lớn nhất của biểu thức \(F=4x-x^2+1\) là 5 khi x=2