1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

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

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

1, x(x-1)=2(x-1)

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

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

<=>x=2 hoặc x=1 

vậy ...

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

<=>(x+2)(2x-3)=(x-2)(x+2)

<=> (x+2)(2x-3)-(x-2)(x+2)=0

<=> (x+2)(2x-3-x+2)=0 

<=> x=-2 hoặc x=1

vây... 

3,x^2 +3x +2=0 

<=> x^2 +x+2x+2=0 

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

<=> x=-2 hoặc x=-1 

vậy ...

5, x^3+x^2-12x =0

<=> x(x^2+x-12)=0

<=>x(x^2-3x+4x-12)=0

<=>x(x+4)(x-3)=0 

<=> x=0 hoặc x=-4 hoặc x=3

vậy ... 

Câu 1:

\(M=x^2-3x+5\)

\(M=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}\)

\(M=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

            Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

    Vậy Min M = 11/4 khi x=3/2

b)\(N=2x^2+3x\)

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

\(N=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)-\frac{9}{8}\)

\(N=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

              Dấu = xảy ra khi \(x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)

                       Vậy MIn N = -9/8 khi x=-3/4

c)Tự làm nha

Ta có : x² - 3x + 5 

= x² - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\) + \(\frac{11}{4}\)

= \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\in R\)

Nên : \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\) \(\ge\frac{11}{4}\forall x\in R\)

Vậy GTNN của biểu thức là : \(\frac{11}{4}\) khi \(x=\frac{3}{2}\)

Câu 2:

a)\(A=-x^2-5x+3\)

\(A=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{37}{4}\)

\(A=\frac{37}{4}-\left(x+\frac{5}{2}\right)^2\le\frac{37}{4}\)

            Dấu = xảy ra khi \(x+\frac{5}{2}=0\Rightarrow x=-\frac{5}{2}\)

                      Vậy Max A = 37/4 khi x=-5/2

b)\(B=-2x^2+3x\)

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

\(B=-2\left(x^2-2.\frac{3}{4}+\frac{9}{16}\right)+\frac{9}{8}\)

\(B=\frac{9}{8}-2\left(x-\frac{3}{4}\right)^2\le\frac{9}{8}\)

         Dấu = xảy ra khi \(x-\frac{3}{4}=0\Rightarrow x=\frac{3}{4}\)

                    Vậy Max B=9/8 khi x=3/4

a: 3x-5>15-x

=>4x>20

hay x>5

b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)

=>3x²+x>3x²-12

=>x>-12

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

2. a. \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18\)

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

Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-2\right)^2-18\ge-18\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy minA = - 18 <=> x = 2

b. \(B=9x-3x^2=-3\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)

\(=-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\le\frac{27}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Vậy maxB = 27/4 <=> x = 3/2

Sửa đề:x³-3x²-4x+12

a,x³-3x²-4x+12

=(x³-3x²)-(4x+12)

=x²(x-3)-4(x-3)

=(x²-4)(x-3)

b,x⁴- 5x² +4

x⁴-4x²-x²+4

(x⁴-x²)-(4x²+4)

x²(x²-1)-4(x²-1)

(x²-4)(x²-1)

Bài 1.

a) x³ - 3x² - 4x + 12 ( mạn phép sửa 13 thành 12, chứ để 13 là không phân tích được :> )

= x²( x - 3 ) - 4( x - 3 )

= ( x - 3 )( x² - 4 )

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

b) x⁴ - 5x² + 4

Đặt t = x²

Đa thức <=> t² - 5t + 4

= t² - t - 4t + 4

= t( t - 1 ) - 4( t - 1 )

= ( t - 1 )( t - 4 )

= ( x² - 1 )( x² - 4 )

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

c) ( x + y + z )³ - x³ - y³ - z³

= ( x + y + z )³ - ( x³ + y³ + z³ )

= ( x + y + z )³ - [ ( x + y + z )³ - 3( x + y )( y + z )( z + x ) ] ( chỗ này bạn xem HĐT tổng ba lập phương nhé )

= ( x + y + z )³ - ( x + y + z )³ + 3( x + y )( y + z )( z + x )

= 3( x + y )( y + z )( z + x )

d) 45 + x³ - 5x² - 9x 

= ( x³ - 5x² ) - ( 9x - 45 )

= x²( x - 5 ) - 9( x - 5 )

= ( x - 5 )( x² - 9 )

= ( x - 5 )( x - 3 )( x + 3 )

e) x⁴ - 2x³ + 3x² - 2x - 3 ( sửa -3x³ -> 3x² )

= x⁴ - x³ - x³ + 3x² - x² + x² - 3x + x - 3

= ( x⁴ - x³ + 3x² ) - ( x³ - x² + 3x ) - ( x² - x + 3 )

= x²( x² - x + 3 ) - x( x² - x + 3 ) - 1( x² - x + 3 )

= ( x² - x - 1 )( x² - x + 3 )

Bài 2.

A = 2x² - 8x - 10

= 2( x² - 4x + 4 ) - 18

= 2( x - 2 )² - 18 

2( x - 2 )² ≥ 0 ∀ x => 2( x - 2 )² - 18 ≥ -18

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MinA = -18 <=> x = 2

B = 9x - 3x²

= -3( x² - 3x + 9/4 ) + 27/4

= -3( x - 3/2 )² + 27/4

-3( x - 3/2 )² ≤ 0 ∀ x => -3( x - 3/2 )² + 27/4 ≤ 27/4

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MaxB = 27/4 <=> x = 3/2

Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ 

1)\(\sqrt{4x^2+12x+9}=2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

\(\)

2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\)       ĐK:\(x\ge-1\)

\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)