Bài 1 : tìm x
A) 3\2 —2\3× x = 7/—9
b)|x|— 5\2=3\4
C) |x—1/2| + 5\3 = 7\3
D) 34\25× x2= 18\25
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Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)
a: =>x+7/4=6:2/3=9
=>x=29/4
b: =>x:5/3=7/5
=>x=7/5*5/3=7/3
c:=>x+1/6=5/3
=>x=10/6-1/6=3/2
d: =>x+4/5=4/5+3/7+3/5
=>x=3/7+3/5=36/35
e: =>x/35=4/5-5/7=3/35
=>x=3
f: =>13/28+x=1/2
=>x=1/28
g: =>1/3-x=1/9
=>x=2/9
l) (x + 9) . (x2 – 25) = 0
<=> (x + 9) . (x – 5) . (x + 5) = 0
<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{-9,5,-5\right\}\)
e) |x - 4 |< 7
<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
Vậy S = \(\left\{11;-3\right\}\)
I,(x+9).(x^2-25)=0
tương đương:x+9=0
x^2-25=0
tương đương : x=-9
x=5
e,\(\left|x-4\right|\)=7
tương đương x-4=4
x-4=-4
tương đương :x=0
x=-8
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
` 8/23 . 46/24 =1/3 .x`
`=>8/23 . 23/12 =1/3 . x`
`=> 1/3 . x=2/3`
`=>x=2/3 : 1/3`
`=>x=2/3 . 3`
`=> x= 6/3`
`=>x=2`
`----`
`1/5 : x= 1/5-1/7`
`=>1/5 : x= 7/35 - 5/35`
`=> 1/5 :x= 2/35`
`=>x= 1/5 : 2/35`
`=>x=1/5 . 35/2`
`=>x=7/2`
`----`
`4/9 - (x-1/2)^2 =1/3`
`=> (x-1/2)^2 =4/9-1/3`
`=> (x-1/2)^2 =4/9- 3/9`
`=> (x-1/2)^2 =1/9`
`=> (x-1/2)^2 = (+- 1/3)^2`
`@ TH1`
`x-1/2=1/3`
`=>x=1/3+1/2`
`=>x= 2/6 + 3/6`
``=>x= 5/6`
`@ TH2`
`x-1/2=-1/3`
`=>x=-1/3 +1/2`
`=>x= -2/6 + 3/6`
`=>x=1/6`
`----`
`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`
`=> 3,2 . x-22/15 : 11/3 = 7/10`
`=> 3,2 . x-22/15 = 7/10 . 11/3`
`=> 3,2 . x-22/15 =77/30`
`=> 3,2 .x= 77/30 + 22/15`
`=> 3,2 .x=121/30`
`=>x= 121/30. 5/16`
`=>x= 121/96`
Bài 1:
a: x/-2=-18/x
=>x2=36
=>x=6 hoặc x=-6
b: x/2+x/5=17/10
=>7/10x=17/10
hay x=17/7
a)2/3x=41/18
x=41/12
b)lxl=13/4
=>x=+-13/4
c)lx-1/2l=4
=>x+1/2=+-4
th1x+1/2=-4
=>x=-9/2
th2 x+1/2=4
=>x=3.5
d)x^2=9/17
câu này mk bó tay