tìm x biết:
x+2x+3x+4x+.....+9x=1000-25.4
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Em tách ra thành:
x(1+3+5+...+2021)-x(2+4+...+2020)=2022.
Sau đó giải bình thường.
Chúc em học tốt!
f: Ta có: \(x\left(2x-9\right)-4x+18=0\)
\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)
g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)
\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)
f. x(2x - 9) - 4x + 18 = 0
<=> x(2x - 9) - 2(2x - 9) = 0
<=> (x - 2)(2x - 9) = 0
<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)
g. 4x(x - 1000) - x + 1000 = 0
<=> 4x(x - 1000) - (x - 1000) = 0
<=> (4x - 1)(x - 1000) = 0
<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)
h. 2x(x - 4) - 6x2(-x + 4) = 0
<=> 2x(x - 4) + 6x2(x - 4) = 0
<=> (2x + 6x2)(x - 4) = 0
<=> 2x(1 + 3x)(x - 4) = 0
<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)
i. 2x(x - 3) + x2 - 9 = 0
<=> 2x(x - 3) + (x - 3)(x + 3) = 0
<=> (2x + x + 3)(x - 3) = 0
<=> (3x + 3)(x + 3) = 0
<=> \(\left[{}\begin{matrix}3x+3=0\\x+3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
j. 9x - 6x2 + x3 = 0
<=> x(9 - 6x + x2) = 0
<=> x(3 - x)2 = 0
<=> \(\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
$ 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) ĐK : x khác 2/3 ; x khác 0
\(\frac{x+5}{3x-2}=\frac{A}{x\left(3x-2\right)}\)
\(\Leftrightarrow\frac{x\left(x+5\right)}{x\left(3x-2\right)}=\frac{A}{x\left(3x-2\right)}\)
\(\Leftrightarrow A=x^2+5x\)
b) \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\frac{2\left(2-x\right)}{\left(x+2\right)}\)
\(=\frac{-5}{2}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`
`=> 3x (12x-4) - 3*3x (4x - 3) = 30`
`=> 3x [12x - 4 - 3(4x-3)] = 30`
`=> 3x (12x - 4 - 12x + 9) = 30`
`=> 3x (-4+9)=30`
`=> 3x*5=30`
`=> 3x=6`
`=> x=2`
Vậy, `x=2`
`b)`
`x( 5 - 2x) + 2x( x - 1)`
`=> x(5-2x) + 2x^2 - 2x=15`
`=> 5x - 2x^2 + 2x^2 - 2x =15`
`=> 3x = 15`
`=> x=5`
Vậy, `x=5.`
a: =>36x^2-12x-36x^2+27x=30
=>15x=30
=>x=2
b: =>5x-2x^2+2x^2-2x=15
=>3x=15
=>x=5
`# \text {DNamNgV}`
`a)`
`(2x + 1)(3x - 2) + 9x^2 = 4`
`<=> 2x(3x - 2) + 3x - 2 + 9x^2 = 4`
`<=> 6x^2 - 4x + 3x - 2 + 9x^2 = 4`
`<=> 15x^2 - x - 2 = 4`
`<=> 15x^2 - x - 2 - 4 = 0`
`<=> 15x^2 - x - 6 = 0`
`<=> 15x^2 + 9x - 10x - 6 = 0`
`<=> (15x^2 + 9x) - (10x + 6) = 0`
`<=> 3x(5x + 3) - 2(5x + 3) = 0`
`<=> (3x - 2)(5x + 3) = 0`
`<=>` TH1: `3x - 2 = 0`
`<=> 3x = 2`
`<=> x = 2/3`
TH2: `5x + 3 = 0`
`<=> 5x = -3`
`<=> x = -3/5`
Vậy, `x \in {-3/5; 2/3}.`
`b)`
`x^3 - 4x = 0`
`<=> x(x^2 - 4) = 0`
`<=> x*(x^2 - 2^2) = 0`
`<=> x(x - 2)(x + 2) = 0`
`<=>` TH1: `x = 0`
TH2 `x - 2 = 0`
`<=> x = 2`
TH3: `x + 2 = 0`
`<=> x = -2`
Vậy, `x \in {+-2; 0}.`
Ta có;
x+2x+3x+4x+...+9x=1000-100
=>x+2x+3x+4x+...+9x=900
=>(1+2+3+4+....+9)x=900
=>[(1+9).9:2]x=900
=>45x=900
x=900:45=20
cho tớ đúng nha
Ta có;
x+2x+3x+4x+...+9x=1000-100
=>x+2x+3x+4x+...+9x=900
=>(1+2+3+4+....+9)x=900
=>45x=900
x=900:45=20