Tìm x
a, 2 x ( X - 1/3 ) - ( X - 1/2 ) = 1/2 x X
b, -3 x ( X - 1/4 ) - 1/3 x ( X + 1/2 ) = X
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` 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`
`@` ` \text {Ans}`
`\downarrow`
`a,`
`1/4+3/4*x=3/2-x`
`=> 1/4 + 3/4x - 3/2 + x = 0`
`=> (1/4 - 3/2) + (3/4x + x) = 0`
`=> -5/4 + 7/4x = 0`
`=> 7/4x = 5/4`
`=> x = 5/4 \div 7/4`
`=> x = 5/7`
Vậy, `x=5/7`
`b,`
`3/5*x-1/4=1/10*x-1/2`
`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`
`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`
`=> 1/2x + 1/4 = 0`
`=> 1/2x = -1/4`
`=> x = -1/4 \div 1/2`
`=> x = -1/2`
Vậy, `x=-1/2`
`c,`
`3x-3/5=x-1/4`
`=> 3x - 3/5 - x + 1/4 = 0`
`=> (3x - x) - (3/5 - 1/4) = 0`
`=> 2x - 7/20 = 0`
`=> 2x = 0,35`
`=> x = 0,35 \div 2`
`=> x = 7/40`
Vậy, `x=7/40`
`d,`
`3/2*x-2/5=1/3*x-1/4`
`=> 3/2x - 2/5 - 1/3x + 1/4 = 0`
`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`
`=> 7/6x - 3/20 = 0`
`=> 7/6x = 3/20`
`=> x = 3/20 \div 7/6`
`=> x = 9/70`
Vậy, `x=9/70`
`@` `\text {Kaizuu lv uuu}`
b)(x+3)2-(x-4)(x+8)=1
\(\Rightarrow\)x2+6x+9-(x2+8x-4x-32)=1
⇒x2+6x+9-x2-8x+4x+32=1
⇒2x+41=1
\(\Rightarrow\)2x+41-1=0
\(\Rightarrow\)2x+40=0
⇒2x=-40
\(\Rightarrow\)x=\(\dfrac{-40}{2}\)
⇒x=-20
Lời giải:
a. Do $|x+1|+|x+2|\geq 0$ với mọi $x$ theo tính chất trị tuyệt đối
$\Rightarrow x\geq 0$
$\Rightarrow x+1, x+2>0\Rightarrow |x+1|=x+1; |x+2|=x+2$. Khi đó:
$(x+1)+(x+2)=x$
$\Leftrightarrow x=-3$ (loại do $x\geq 0$)
Vậy không tồn tại $x$ thỏa mãn
b. Tương tự phần a:
$|x+1|+|x+2|+|x+3|\geq 0\Rightarrow 2x\geq 0\Rightarrow x\geq 0$
$\Rightarrow x+1, x+2, x+3>0$
$\Rightarrow |x+1|=x+1; |x+2|=x+2; |x+3|=x+3$. Khi đó:
$(x+1)+(x+2)+(x+3)=2x$
$\Leftrightarrow x=-6< 0$ (loại)
Vậy không tồn tại $x$ thỏa mãn.
c.
$|x+1|+|x+2|+|x+3|+|x+4|\geq 0$
$\Rightarrow 3x\geq 0\Rightarrow x\geq 0$
$\Rightarrow x+1,x+2, x+3, x+4>0$
$\Rightarrow |x+1|=x+1, |x+2|=x+2, |x+3|=x+3, |x+4|=x+4$. Khi đó:
$(x+1)+(x+2)+(x+3)+(x+4)=3x$
$4x+10=3x$
$x=-10< 0$ (loại vì $x\geq 0$)
Vậy không tồn tại $x$ thỏa mãn
d.
$|x+1|+|x+2|+|x+3|+|x+4|+|x+5|\geq 0$
$\Rightarrow 4x\geq 0\Rightarrow x\geq 0\Rightarrow x+1,x+2,x+3,x+4,x+5>0$
$\Rightarrow |x+1|=x+1, |x+2|=x+2, |x+3|=x+3, |x+4|=x+4, |x+5|=x+5$. Khi đó:
$(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=4x$
$5x+15=4x$
$x=-15< 0$ (loại vì $x\geq 0$)
Vậy không tồn tại $x$ thỏa đề.
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
Bài 3:
b: \(x^2+2x+1=\left(x+1\right)^2\)
c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20
a) 2.(x-1/3) - (x-1/2) = 1/2.x
2.x - 2/3 - x + 1/2 = 1/2.x
=> 2.x-x - 1/2.x = 2/3 -1/2
1/2.x = 1/6
x = 1/3
bài b bn làm tương tự nha