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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
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
23 tháng 7 2018
a) \(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)
\(\Rightarrow6x^2+2x+12x-6x^2=7\)
\(\Rightarrow14x=7\Rightarrow x=\frac{1}{2}\)
b) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(-80x=-480\)
x = 6
c) \(\left(3x+2\right).\left(2x+9\right)-\left(x+2\right).\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Rightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x+1-x+6\) ( chỗ này bn tự phân tích ik nha, mk chỉ đưa ra kp sau khi phân tích thôi, ko thì viết ra dài lắm)
\(\Rightarrow18x+16=7\)
18x = -9
x = -2
18x =
31 tháng 7 2022
a: \(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
=>14x=7
hay x=1/2
b: \(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
=>-56x+156=24x-324
=>-80x=-480
hay x=6
c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2
20 tháng 12 2021
a: \(\Leftrightarrow-15x+10=-7x+14\)
=>-8x=4
hay x=-1/2
20 tháng 12 2021
\(a,\dfrac{2-3x}{x-2}=-\dfrac{7}{5}\left(x\ne2\right)\\ \Leftrightarrow14-7x=10-15x\\ \Leftrightarrow8x=-4\Leftrightarrow x=-2\left(tm\right)\\ c,\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{5}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{2\cdot2+5\cdot3-4}=\dfrac{45}{15}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=6\\y-2=15\\z-3=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=17\\z=15\end{matrix}\right.\\ d,\Leftrightarrow\dfrac{x}{1}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{6x+7y+8z}{24+84+120}=\dfrac{456}{228}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x=8\\y=24\\z=30\end{matrix}\right.\)
25 tháng 5 2022
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
8 tháng 7 2017
Ta có : |2x - 1| + 1 = x
=> |2x - 1| = x - 1
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x-1\\2x-1=1-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-x=-1+1\\2x+x=1+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
a) Ix+7I -x =7
Ix+7I = 7+x = x+7
Vậy có vô số x thỏa mản điều kiện
b) Xét 3 trường hợp
Th1: 3x-7> 0 => I3x-7I = 3x-7
=> 3x-7=2x+1
=> 3x-2x= 1+7
=> x= 8.
Th2 3x-7 < 0 => I3x-7I = -(3x-7)
=> -(3x-7) = 2x+1
=> -3x+7 = 2x +1
=> 7-1 = 2x+3x
=> 6 = 5x => x=6/5
Th3: I3x-7I= 0 => 3x-7 =0
=> 2x+1= 0
=> 2x= -1 = x= -1/2
Vậy x= -1/2 hoặc 6/5 hoặc 8
c) Bạn cũng xét 3 trường hợp như vậy nhé