-10+3|.x-2|=-1
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Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
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Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
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\(\frac{1}{2}.\frac{9}{10}+\frac{1}{2}.\frac{8}{10}+\frac{1}{2}.\frac{3}{10}=\frac{1}{2}.\left(\frac{9}{10}+\frac{8}{10}+\frac{3}{10}\right)=\frac{1}{2}.2=1\)
Dấu chấm là dấu nhân nhé
= \(\frac{1}{2}\times\left(\frac{9}{10}+\frac{8}{10}+\frac{3}{10}\right)=\frac{1}{2}\times\frac{20}{10}=\frac{1}{2}\times2=1\)
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`2 1/2 xx 7/5 + (-9/10) xx 2/3`
`= 5/2 xx 7/10 + (-9/10) xx 2/3`
`= 35/20 + (-18/30)`
`= 7/4 + (-3/5)`
`= 23/20`
__
`x + 45% =2 1/2 - 5/3`
`=> x+ 45/100 = 5/2 - 5/3`
`=>x+ 9/20= 15/6-10/6`
`=> x+9/20 = 5/6`
`=>x= 5/6 - 9/20`
`=>x=23/60`
__
`80% +x=-5/2 +3/4`
`=> 80/100 + x= -10/4 +3/4`
`=> 4/5 + x= -7/4`
`=>x= -7/4-4/5`
`=>x=-51/20`
__
`4/25 -x=-5/2 +(-3/10)`
`=> 4/25 -x= -25/10 +(-3/10)`
`=> 4/25 -x= -28/10`
`=>x= 4/25 -(14/5)`
`=>x= 4/25 + 14/5`
`=>x=74/25`
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a: =>14x+20+5=6x-9-9x
=>14x+25=-3x-9
=>17x=-34
=>x=-2
b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)
=>-28x-30=-50
=>-28x=-20
=>x=20/28=5/7
c: =>2x+x^3-x=x^3+1
=>x=1
d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22
=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22
=>-5x^2+2x-1+2x^2+x+22=0
=>-3x^2+3x+21=0
=>x^2-x-7=0
=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)
-10 + 3|x - 2| = -1
=> 3|x - 2| = -1 - (-10)
=> 3|x - 2| = 9
=> |x - 2| = 9 : 3
=> |x - 2| = 3
=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\)
=> \(\orbr{\begin{cases}x=3+2\\x=-3+2\end{cases}}\)
=> \(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
3!x-2!=-1-(-10) muốn tìm số hạng lấy tổng trừ đi số hạng còn lại
3!x-2!=-1+10=9
!x-2!=9:3 {muốn tìm thừa số lấy tích chia cho thừa số còn lại}
!x-2!=3
\(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\) Quy tắc phá trị tuyệt đối cơ bản