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Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

12 tháng 7 2021

ai giúp mik vs

14 tháng 11 2023

a: \(\left|7-2x\right|+7=2x\)

=>\(\left|2x-7\right|+7=2x\)

=>\(\left|2x-7\right|=2x-7\)

=>2x-7>=0

=>\(x>=\dfrac{7}{2}\)

b: \(\left|1-x\right|=4x+1\)

=>\(\left|x-1\right|=4x+1\)

=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)

=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)

d: \(\left|x-7\right|+2x+5=6\)

=>\(\left|x-7\right|=6-2x-5=-2x+1\)

=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)

e: 3x-|2x-1|=2

=>|2x-1|=3x-2

=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

18 tháng 7 2023

giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

b: =>4/5x-3/2x=1/2+6/5

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

d: =>6x+3/2+4/5=1/2-2x

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

27 tháng 9

         Bài 1:

\(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)

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

-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)

\(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)

  \(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))

 \(x\) = \(\dfrac{3}{14}\)

Vậy \(x=\dfrac{3}{14}\)

 

 

27 tháng 9

Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1

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

         - 5\(x\)    = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\) 

        - 5\(x\)    = \(\dfrac{7}{6}\)

           \(x\)    = \(\dfrac{7}{6}\) : (- 5) 

          \(x\)    = - \(\dfrac{7}{30}\)

Vậy \(x=-\dfrac{7}{30}\)

 

17 tháng 12 2022

a: \(\Leftrightarrow4^{x-5}\cdot17=68\)

=>4^x-5=4

=>x-5=1

=>x=6

b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)

=>|2x-1|=1/3

=>2x-1=1/3 hoặc 2x-1=-1/3

=>x=2/3 hoặc x=1/3

c: =>|2x-2|=|3x+15|

=>3x+15=2x-2 hoặc 3x+15=-2x+2

=>x=-17 hoặc x=-13/5

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

5 tháng 9 2018

a,  \((\frac{3}{7}-\frac{2}{3})\)  .x  =\(\frac{10}{21}\)                                                                   

  \(\frac{-5}{21}\).x=\(\frac{10}{21}\)

x= -2

 Mk chỉ làm 1 phần  các phằn còn lại tương tự

5 tháng 9 2018

Bài rút gọn biểu thức

a) M=|2x-3|+|x-1| với x > 1,5

b) N=|2-x|-3|x+1| với x < -1

c) P=|3x-5|+|x-2|

d) Q=|x-3|-2.|-5x|

`#3107.101107`

`1.`

`a,`

`(2x - 3)^2 = |3 - 2x|`

`=> (2x - 3)^2 = |2x - 3|`

`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)

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

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

Vậy, `x \in {3/2; 2; 1}`

`b,`

`(x - 1)^2 + (2x - 1)^2 = 0`

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

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

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

Vậy, `x \in {1; 1/2}`

`c,`

`5 - x^2 = 1`

`=> x^2 = 4`

`=> x^2 = (+-2)^2`

`=> x = +-2`

Vậy, `x \in {-2; 2}`

`d,`

`x - 2\sqrt{x} = 0`

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

`=> x^2 - 4x = 0`

`=> x(x - 4) = 0`

`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy, `x \in {0; 4}`

`g,`

`(x - 1) + 1/7 = 0`

`=> x - 1 + 1/7 = 0`

`=> x - 6/7 = 0`

`=> x = 6/7`

Vậy, `x = 6/7.`