a: x=-5/11+2/11=-3/11

b: =>x=-3/24+20/24+1/24=18/24=3/4

c: =>5/8-x=1/9+5/4=4/36+45/36=49/36

=>x=5/8-49/36=-53/72

d: =>2/3-x=1/3

=>x=1/3

e: =>1/5:x=12/35

=>x=7/12

\(A=\left(4x^2+4x+1\right)+10=\left(2x+1\right)^2+10\ge10\)

\(A_{min}=10\) khi \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)

\(B=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)

\(B_{min}=-36\) khi \(x^2+5x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(C=\left(x^2-2x+1\right)+\left(y^2-4x+4\right)+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left(x;y\right)=\left(1;2\right)\)

a. \(A=4x^2+4x+11\)

   \(A=\left(4x^2+4x+1\right)+10\)

  \(A=\left(2x+1\right)^2+10\)

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

\(\Rightarrow A_{min}=10\)

Dấu "=" xảy ra khi \(\left(2x+1\right)^2=0\)

                            \(\Leftrightarrow2x+1=0\Leftrightarrow x=-\dfrac{1}{2}\)

c.\(C=x^2-2x+y^2-4y+7\)

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

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

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

\(\Rightarrow C_{min}=2\)

Dấu "=" xảy ra khi\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\)

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

3. Tìm giá trị nhỏ nhất của các biểu thứca. A = 4x2  4x 11b. B = (x - 1) (x 2) (x 3) (x 6)c. C = x2 - 2x y2 - 4y 7Ai nha... - Hoc24

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

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

\(\Leftrightarrow x=2\)

b: Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+2\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x^3-3x^2-2x+3x^2=0\)

\(\Leftrightarrow2x=64\)

hay x=32

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

⇔  \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

<=> (x-1)(3x+1)=0

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

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

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

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

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

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

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x²  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

<=> x(x-1)+3(x-1)=0

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

b) \(\text{3x (x-1) + x - 1 = 0}\)

\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

c) \(\text{2(x+3) - x ² - 3x = 0}\)

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Rightarrow\left(2-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d) \(\text{x(x - 2) + 3x - 6 = 0}\)

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

e)

\(\text{4x ² - 4x +1 = 0}\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow2x-1=0\\ \Rightarrow x=0,5\)

f) \(\text{x +5x ² = 0}\)

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

viết lại câu g đi bạn

a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)

=>-8x+3=11

=>-8x=8

hay x=-1

b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)

=>-11x=10

hay x=-10/11

Bài 2:

a: 4x(x-3)+6(3-x)=0

=>4x(x-3)-6(x-3)=0

=>(x-3)(4x-6)=0

=>\(\left[{}\begin{matrix}x-3=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\)

b: \(x^3-x\left(x+1\right)\left(x-1\right)=14\)

=>\(x^3-x\left(x^2-1\right)=14\)

=>\(x^3-x^3+x=14\)

=>x=14

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

=>\(\left(x^2-x\right)^2+2\left(x^2-x\right)-8=0\)

=>\(\left(x^2-x\right)^2+4\left(x^2-x\right)-2\left(x^2-x\right)-8=0\)

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

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

=>\(\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\right)\left(x-2\right)\left(x+1\right)=0\)

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

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

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

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

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

Lời giải:

a.

$\frac{2}{3}+\frac{1}{3}:3\times x=20\text{%}$
$\frac{2}{3}+\frac{1}{9}\times x=\frac{1}{5}$

$\frac{1}{9}\times x=\frac{1}{5}-\frac{2}{3}=\frac{-7}{15}$

$x=\frac{-7}{15}: \frac{1}{9}=\frac{-21}{5}$
b.

$\frac{3-x}{5-x}=\frac{6}{11}$
$\Rightarrow 6(5-x)=11(3-x)$

$\Rightarrow 30-6x=33-11x$

$\Rightarrow 5x=3$

$\Rightarrow x=\frac{3}{5}$

x2x2 là sao bn