1) Tìm x biết:5.(x+1)=25
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`(1/(1.3)+1/(3.5)+.......+1/(23.25))xx((x+1)+(x+3)+(x+5)+.....+(x+23))=144`
`(2/(1.3)+2/(3.5)+.......+2/(23.25))xx[(x+x+....+x)+(1+3+5+...+23)]=288`
`(1-1/3+1/3-1/5+.....+1/23-1/25)xx(12x+(24.12)/2)=288`
`(1-1/25)xx(12x+12.12)=288`
`24/25xx[12(x+12)]=288`
`24/25xx(x+12)=28`
`x+12=28:24/25=50`
`x=50-12=38`
Vậy `x=38`
Lời giải:
a.
$x=\frac{7}{25}+\frac{-1}{5}=\frac{7}{25}+\frac{-5}{25}=\frac{7-5}{25}=\frac{2}{25}$
b.
$x=\frac{5}{11}+\frac{4}{-9}=\frac{5}{11}-\frac{4}{9}=\frac{45}{99}-\frac{44}{99}=\frac{1}{99}$
c.
$\frac{x}{-1}=\frac{-1}{3}-\frac{5}{9}=\frac{-3}{9}-\frac{5}{9}=\frac{-8}{9}$
$x=(-1).\frac{-8}{9}=\frac{8}{9}$
a) \(A\left(x\right)=x^2-10x+25\)
\(\Rightarrow A\left(x\right)=\left(x-5\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}A\left(0\right)=\left(0-5\right)^2=25\\A\left(-1\right)=\left(-1-5\right)^2=36\end{matrix}\right.\)
b) \(A\left(x\right)+B\left(x\right)=6x^2-5x+25\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-\left(x^2-10x+25\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-x^2+10x-25\)
\(\Rightarrow B\left(x\right)=5x^2+5x\)
\(\Rightarrow B\left(x\right)=5x\left(x+1\right)\)
c) \(A\left(x\right)=\left(x-5\right)C\left(x\right)\)
\(\Rightarrow C\left(x\right)=\dfrac{\left(x-5\right)^2}{x-5}=x-5\left(x\ne5\right)\)
d) Nghiệm của B(x)
\(\Leftrightarrow B=0\)
\(\Leftrightarrow5x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của B(x)
\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=-25\)
\(\Leftrightarrow3x^2-6x+3-3x^2-13x+10=-25\)
\(\Leftrightarrow-19x=-38\Leftrightarrow x=2\)
\(\Rightarrow3x^2-6x+3+2x-3x^2+10-15x=-25\\ \Rightarrow-19x=-38\\ \Rightarrow x=2\)
1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)
\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)
2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)
3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)
1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)
\(\frac{3}{5}-\frac{1}{4}:x=\frac{25}{100}\)
\(\frac{3}{5}-\frac{1}{4}:x=\frac{1}{4}\)
\(\frac{1}{4}:x=\frac{3}{5}-\frac{1}{4}\)
\(\frac{1}{4}:x=\frac{7}{20}\)
\(x=\frac{1}{4}:\frac{7}{20}\)
\(x=\frac{5}{7}\)
bg:
3/5-1/4:x =25/100
1/4:x =3/5-25/100
1/4:x =7/20
x = 1/4:7/20
x =5/7
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?
5.(x+1)=25
x+1 = 25 : 5
x+1 = 5
x = 5-1
x = 4
Vay x=4