tìm x, biết 2/9x 5x+1/2 -1/18=5/36
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a, \(\frac{5}{7}.3x-\frac{8}{5}=\frac{9}{35}\)
=> \(\frac{15}{7}x=\frac{9}{35}+\frac{8}{5}\)=> \(\frac{15}{7}x=\frac{9}{35}+\frac{56}{35}\)
=> \(\frac{15}{7}x=\frac{65}{35}=\frac{13}{7}\)=> \(x=\frac{13}{7}:\frac{15}{7}=\frac{13}{15}\)
vậy \(x=\frac{13}{15}\)
b, \(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)
=> \(\frac{10}{9}x+\frac{1}{2}=\frac{5}{36}+\frac{1}{18}\)=\(\frac{5}{36}+\frac{2}{36}=\frac{7}{36}\)
=> \(\frac{10}{9}x=\frac{7}{36}-\frac{1}{2}\)=\(\frac{7}{36}-\frac{18}{36}\)=\(\frac{-11}{36}\)=> \(x=\frac{-11}{36}:\frac{10}{9}\)=\(\frac{-11}{36}.\frac{9}{10}\)=\(\frac{-11}{40}\)
vậy x=\(\frac{-11}{40}\)
a, ta có : \(\frac{5}{7}.\frac{3x-8}{5}=\frac{9}{35}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}:\frac{5}{7}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}.\frac{7}{5}\)
\(\Leftrightarrow\frac{3x-8}{5}=\frac{9}{5.5}\Leftrightarrow3x-8=\frac{9}{25}.5\Leftrightarrow3x-8=\frac{9.5}{25}\)
\(\Leftrightarrow3x-8=\frac{9}{5}\Leftrightarrow3x-8=\frac{9}{5}+8\Leftrightarrow3x=\frac{9+8.5}{5}\)
\(\Leftrightarrow3x=\frac{49}{5}\Leftrightarrow x=\frac{49}{5}:3\Leftrightarrow x=\frac{49}{5}.\frac{1}{3}=\frac{49}{15}\)
~ Vậy, ta tìm được \(x=\frac{49}{15}\)
b, Ta có : \(\frac{2}{9}.\frac{5x+1}{2}-\frac{1}{18}=\frac{5}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5}{36}+\frac{1}{18}\)
\(\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5+2.1}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{7}{36}\)
\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}:\frac{2}{9}\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}.\frac{9}{2}\Leftrightarrow\frac{5x+1}{2}=\frac{7.9}{4.9.2}\)
\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{8}\Leftrightarrow5x+1=\frac{7}{8}.2\Leftrightarrow5x+1=\frac{7.2}{8}\)
\(\Leftrightarrow5x+1=\frac{7}{4}\Leftrightarrow5x=\frac{7}{4}-1\Leftrightarrow5x=\frac{7-1.4}{4}\)
\(\Leftrightarrow5x=\frac{3}{4}\Leftrightarrow x=\frac{3}{4}:5\Leftrightarrow x=\frac{3}{4}.\frac{1}{5}\Leftrightarrow x=\frac{3}{20}\)
~ Vậy, ta tìm được \(x=\frac{3}{20}\)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
\(a,12x=4x-30\Leftrightarrow8x=-30\Leftrightarrow x=-\dfrac{15}{4}\)
\(b,2x-5=x-1\Leftrightarrow2x-x=-1+5\Leftrightarrow x=4\)
\(c,2-5x=5x-10\Leftrightarrow-10x=-12\Leftrightarrow x=\dfrac{6}{5}\)
\(d,9x-6=1x-5\Leftrightarrow8x=1\Leftrightarrow x=\dfrac{1}{8}\)
\(e,2x-5=2x-1\Leftrightarrow2x-2x=-1+5\Leftrightarrow0x=4\) (Vô lí)\(\Rightarrow x\in\varnothing\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
Bài 1.
a) ( 7x - 3 )2 - 5x( 9x + 2 ) - 4x2 = 18
<=> 49x2 - 42x + 9 - 45x2 - 10x - 4x2 = 18
<=> -52x + 9 = 18
<=> -52x = 9
<=> x = -9/52
b) ( x - 7 )2 - 9( x + 4 )2 = 0
<=> x2 - 14x + 49 - 9( x2 + 8x + 16 ) = 0
<=> x2 - 14x + 49 - 9x2 - 72x - 144 = 0
<=> -8x2 - 86x - 95 = 0
<=> -8x2 - 10x - 76x - 95 = 0
<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0
<=> ( x + 5/4 )( -8x - 76 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)
c) ( 2x + 1 )2 + ( 4x - 1 )( x + 5 ) = 36
<=> 4x2 + 4x + 1 + 4x2 + 19x - 5 = 36
<=> 8x2 + 23x - 4 - 36 = 0
<=> 8x2 + 23x - 40 = 0
=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))
Bài 2.
a) x2 - 12x + 39 = ( x2 - 12x + 36 ) + 3 = ( x - 6 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
b) 17 - 8x + x2 = ( x2 - 8x + 16 ) + 1 = ( x - 4 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
c) -x2 + 6x - 11 = -( x2 - 6x + 9 ) - 2 = -( x - 3 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
d) -x2 + 18x - 83 = -( x2 - 18x + 81 ) - 2 = -( x - 9 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
`1,`
`538 - x = 275`
`\Rightarrow x = 538 - 275`
`\Rightarrow x = 263`
Vậy, `x = 263`
`2,`
`45 - 9x = 18`
`\Rightarrow 9x = 45 - 18`
`\Rightarrow 9x = 27`
`\Rightarrow x = 27 \div 9`
`\Rightarrow x = 3`
Vậy, `x = 3`
`3,`
`(5x - 9) \div 3 = 12`
`\Rightarrow 5x - 9 = 12. 3`
`\Rightarrow 5x - 9 = 36`
`\Rightarrow 5x = 36 + 9`
`\Rightarrow 5x = 45`
`\Rightarrow x = 45 \div 5`
`\Rightarrow x = 9`
Vậy, `x = 9.`
1) \(538-x=275\)
\(x=538-275\)
\(x=263\)
2) \(45-9x=18\)
\(9x=27\)
\(x=3\)
3) \(\left(5x-9\right)\div3=12\)
\(\left(5x-9\right)=12\times3\)
\(5x-9=36\)
\(5x=45\)
\(x=9\)
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
\(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)
\(\frac{2}{9}.5x=\frac{5}{36}+\frac{1}{18}-\frac{1}{2}\)
\(\frac{2}{9}.5x=\frac{-11}{36}\)
\(5x=\frac{-11}{36}:\frac{2}{9}\)
\(5x=\frac{-11}{8}\)
\(x=\frac{-11}{8}:5\)
\(x=\frac{-11}{40}\)
Chú ý dấu chấm là dâu nhân nha
nhầm rồi 5x+1/2