\(\dfrac{1}{5}\)x\(\dfrac{7}{6}\)+\(\dfrac{1}{5}\)x\(\dfrac{5}{16}\)+\(\dfrac{1}{5}\)x\(\dfrac{3}{16}\)
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a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)
b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)
Bài này có cần phải tính nhanh ko vậy bn?
Nếu ko thì lấy máy tính mà tính cũng đc mà
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
a) ĐKXĐ: \(x\notin\left\{-3;2;-1;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{2}{\left(x+3\right)\left(x+1\right)}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}-\dfrac{2\left(x-2\right)}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{5x+5-2x+4}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{-3}{2x-1}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)
\(\Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x-2\right)}=\dfrac{3}{1-2x}\)
Suy ra: \(\left(x+1\right)\left(x-2\right)=1-2x\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Delta=1^2-4\cdot1\cdot\left(-3\right)=13\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{13}}{2}\left(nhận\right)\\x_2=\dfrac{-1+\sqrt{13}}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-1-\sqrt{13}}{2};\dfrac{-1+\sqrt{13}}{2}\right\}\)
Lớp 8 nên chưa học biệt thức delta
Ta có: \(x^2+x-3=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{13}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{13}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{13}-1}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
Đề bài sai, pt này ko giải được
Đề đúng: \(\dfrac{4x^2+16}{x^2+6}=...\)
Mẫu số bên trái thừa mất số 1
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
quy đồng ra nhé
( 7/6 + 5/16 + 3/16) x 1/5
= ( 7/6 + 1/2) x 1/5
= ( 7/6 + 3/6) x 1/5
= 10/6 x 1/5
= 5/3 x 1/5
=1/3