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AH
Akai Haruma
Giáo viên
30 tháng 1

Lời giải:

$\frac{x-1}{x+3}=\frac{x-2}{x+4}$ (điều kiện: $x\neq -3; -4$)

$\Rightarrow (x-1)(x+4)=(x-2)(x+3)$

$\Rightarrow x^2+3x-4=x^2+x-6$

$\Rightarrow 2x=-2$

$\Rightarrow x=-1$

AH
Akai Haruma
Giáo viên
22 tháng 6 2023

Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.

f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)

=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0

=>6x-24=0

=>x=4

e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2

=>-5x^2-2x+16+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1

=>-16x-34=x-1

=>-17x=33

=>x=-33/17

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6

=>4x^2+16x-20-4x^2-10x+4=0

=>6x=16

=>x=8/3

3 tháng 8 2021

4,  Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)

 xét x \(\ge\) \(-\frac{1}{5}\)

 Ta Có  Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\)  = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\)\(\frac{27}{35}\)   (1)

xét x \(< -\frac{1}{5}\)

Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x  + \(\frac{13}{35}\)

với x \(< -\frac{1}{5}\) 

=> -2x \(>\) \(\frac{2}{5}\) 

=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)

Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)

5 ,  D = |x| + |8-x| 

D = |x| + |8-x| \(\ge\) |x+8-x|  = |8| = 8

Dấu ''='' xảy ra khi   x(8-x) \(\ge\) 0  <=> 0\(\le\)x\(\le\) 8 

Vậy MinD = 8 khi \(0\le x\le8\) 

6,L=  |x - 2012| + |2011 - x| 

L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x |  = |-1| = 1 

Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0  

3 tháng 8 2021

làm nốt câu 6 nãy ấn nhầm 

<=> 2011\(\le\) x \(\le\) 2012

Vậy MinL = 1 khi \(2011\le x\le2012\) 

7 , E = | x- \(\frac{2006}{2007}\) | + |x-1| 

Ta có :

E = |x-\(\frac{2006}{2007}\) | + |1-x| 

E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x |  = \(\frac{1}{2007}\) 

Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=>  \(\frac{2006}{2007}\le x\le1\) 

Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\) 

8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) | 

Ta có :

F  = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\)   - x | 

F  = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x  |  = \(\frac{1}{2}\) 

Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0    <=>  \(\frac{1}{4}\le x\le\frac{3}{4}\) 

Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)

`#040911`

a,

\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)

Vậy, \(x=-\dfrac{8}{21}\)

b,

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

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

Vậy, \(x\in\left\{-2;3\right\}\)

c,

\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)

Bạn xem lại đề có sai kh nhỉ?

31 tháng 8 2023

c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)

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

=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)

=>\(6x-\dfrac{39}{4}=1\)

=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)

=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)

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

=>\(2x-6=3x+6-x+1\)

=>2x-6=2x+7

=>-6=7(vô lý)

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

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

=>3x-2x=-2x

=>3x=0

=>x=0

d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)

=>\(3x^2-3x-2x-4-2x=x^2-x\)

=>\(3x^2-7x-4-x^2+x=0\)

=>\(2x^2-6x-4=0\)

=>\(x^2-3x-2=0\)

=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)

Nguyễn Trà My

Phần a)

\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(32-3x+13=76-x\)

\(116-3x=76-x\)

\(116-76=3x-x\)

\(46=2x\)

\(x=46\div2\)

\(x=13\)

22 tháng 9 2017

a)  \(3.\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

\(3.\left(\frac{1}{2}-x\right)+x=\frac{7}{6}-\frac{1}{3}\)

\(\Rightarrow\frac{3}{2}-3x+x=\frac{5}{6}\)

\(-3x+x=\frac{5}{6}-\frac{3}{2}\)

\(2x=-\frac{2}{3}\)

\(x=-\frac{2}{3}:2\)

\(x=-\frac{1}{3}\)

`1,`

`f(x)+g(x)=(5x^4+4x^2-2x+7)+(4x^4-2x^3+3x^2+4x-1)`

`= 5x^4+4x^2-2x+7+4x^4-2x^3+3x^2+4x-1`

`=(5x^4+4x^4)-2x^3+(4x^2+4x^2)+(-2x+4x)+(7-1)`

`= 9x^4-2x^3+8x^2+2x+6`

Đề phải là `f(x)-g(x)` chứ nhỉ :v?

`f(x)-g(x)=(5x^4+4x^2-2x+7)-(4x^4-2x^3+3x^2+4x-1)`

`= 5x^4+4x^2-2x+7-4x^4+2x^3-3x^2-4x+1`

`= (5x^4-4x^4)+2x^3+(-2x-4x)+(4x^2-3x^2)+(7+1)`

`= x^4+2x^3-6x+x^2+8`