a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)

\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)

\(\Leftrightarrow9x+3=11\)

\(\Leftrightarrow9x=11-3\)

\(\Leftrightarrow9x=8\)

\(\Leftrightarrow x=\dfrac{8}{9}\)

b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)

\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)

\(\Leftrightarrow-62x+12=-50\)

\(\Leftrightarrow-62x=-50-12\)

\(\Leftrightarrow-62x=-62\)

\(\Leftrightarrow x=\dfrac{-62}{-62}\)

\(\Leftrightarrow x=1\)

a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)

\(x^2+x+3x+3-x^2+5x=11\)

\(x+8x+3=11\)

\(x+8x=8\)

\(x\left(8+1\right)=8\)

\(x=\dfrac{8}{9}\)

b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)

\(-62x+12=-50\)

\(-62x=-62\)

\(x=1\)

x^2 + 6x + 11 = 0

x^2 + 6x + 9 + 2 = 0

(x+3)^2 + 2 = 0

Mà (x+3)^2 + 2 > 0 với mọi x

=> pt vô ngiệm

Vậy .........

Hok tốt

x² + 6x + 11 = 0

x . (x + 6) = 11

* x = 11

* x + 6 = 11

        x = 11 - 6

        x = 5

Vậy x = 11 hoặc x = 5

\(x^2+6x+11=0\)

\(x^2+2.x.3+9+2=0\)

\(x^2+2.x.3+3^2=-2\)

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

Vì \(\left(x+3\right)^2\ge0\forall x\)

Mà\(-2< 0\)

\(\Rightarrow x\in\varnothing\)

\(x^2+6x+11=0\)

\(x\left(x+6\right)=-11\)

Ta có bảng:

Vậy..

hok tốt!!

Hello

=x+x+1+x-2+11=6x+1
=1-2+11-1=-x-x-x+6x
9=3x
x=3

1: Ta có: \(4x^2-36=0\)

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

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

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

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

\(A=\left(x^2-2x+1\right)+9\)

\(A=\left(x-1\right)^2+9\)

Mà  \(\left(x-1\right)^2\ge0\)

\(\Rightarrow A\ge9\)

Dấu "=" xảy ra khi :

\(x-1=0\Leftrightarrow x=1\)

Vậy Min A = 9 khi x = 1

\(B=x^2-5x-7\)

\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)

\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)

Mà  \(\left(x-\frac{5}{2}\right)^2\ge0\)

\(\Rightarrow B\ge-\frac{53}{4}\)

Dấu "=" xảy ra khi :

\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)

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

\(3C=9x^2+9x-15\)

\(3C=\left(9x^2+9x+\frac{9}{4}\right)-\frac{69}{4}\)

\(3C=\left(3x+\frac{3}{2}\right)^2-\frac{69}{4}\)

Mà  \(\left(3x+\frac{3}{2}\right)^2\ge0\)

\(\Rightarrow3C\ge-\frac{69}{4}\)

\(\Leftrightarrow C\ge-\frac{23}{4}\)

Dấu "=" xảy ra khi :

\(3x+\frac{3}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy ...

\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)

Vậy \(x=12\).

(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11

=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11

=>x3−6x2+12x−8−(x3+53)+6x2−11=0

=>12x−144=0

=>x=12

Vậy x=12x=12.

cho tôi đúng đi

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

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

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)

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

4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)

\(=\dfrac{1}{6\left(x^2+x+1\right)}\)

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)

\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)

\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

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

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

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

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

5: \(x^4-6x^2+5\)

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

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

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

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

6: \(9x^2-6x-y^2+2y\)

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

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

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