\(\left(2x-5\right)^2=2x-5\)
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HQ
Hà Quang Minh
Giáo viên
12 tháng 1
Đặt \(y = 2x - 5\).
\(\begin{array}{l}\left[ {8{x^3}{{\left( {2x - 5} \right)}^2} - 6{x^2}{{\left( {2x - 5} \right)}^3} + 10x{{\left( {2x - 5} \right)}^2}} \right]:2x{\left( {2x - 5} \right)^2}\\ = \left( {8{x^3}.{y^2} - 6{x^2}.{y^3} + 10x.{y^2}} \right):2x{y^2}\\ = 8{x^3}.{y^2}:2x{y^2} - 6{x^2}.{y^3}:2x{y^2} + 10x.{y^2}:2x{y^2}\\ = 4{x^2} - 3xy + 5\\ = 4{x^2} - 3x\left( {2x - 5} \right) + 5\\ = 4{x^2} - 6{x^2} + 15x + 5\\ = - 2{x^2} + 15x + 5\end{array}\)
16 tháng 2 2021
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
11 tháng 12 2022
(2x+3)^2-2(2x+3)(2x-5)+(2x-5)^2=x^2+6x+64
=>x^2+6x+64=(2x+3-2x+5)^2
=>x^2+6x+64=64
=>x(x+6)=0
=>x=0 hoặc x=-6
25 tháng 11 2021
\(\Leftrightarrow\left|2x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\1-2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
20 tháng 1
a: ĐKXĐ: \(x\notin\left\{\dfrac{5}{2}\right\}\)
\(\log_32x-5=3\)
=>\(log_3\left(2x-5\right)=log_327\)
=>2x-5=27
=>2x=32
=>x=16(nhận)
b: ĐKXĐ: x<>0
\(\log_4x^2=2\)
=>\(log_4x^2=log_416\)
=>\(x^2=16\)
=>\(\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{5}{2}\right\}\)
\(\log_7\left(3x-1\right)=\log_7\left(2x+5\right)\)
=>3x-1=2x+5
=>x=6(nhận)
d: ĐKXĐ: \(x\notin\left\{1;-1;\dfrac{-1+\sqrt{13}}{4};\dfrac{-1-\sqrt{13}}{4}\right\}\)
\(ln\left(4x^2+2x-3\right)=ln\left(3x^2-3\right)\)
=>\(4x^2+2x-3=3x^2-3\)
=>\(x^2+2x=0\)
=>x(x+2)=0
=>\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\notin\left\{-\dfrac{3}{2};\dfrac{1}{3}\right\}\)
\(log\left(2x+3\right)=log\left(1-3x\right)\)
=>2x+3=1-3x
=>5x=-2
=>\(x=-\dfrac{2}{5}\left(nhận\right)\)
NH
2
13 tháng 8 2015
(2x+3)^2+(2x+5)^2-2(2x+3)(2x+5)
=(2x+3)^2-2(2x+3)(2x+5)+(2x+5)^2
=(2x+3+2x+5)^2
=(4x+8)^2
(2x - 5)2 = 2x - 5
=> (2x - 5)2 - (2x - 5) = 0
=> (2x - 5).(2x - 5 - 1) = 0
=> (2x - 5).(2x - 6) = 0
=> \(\orbr{\begin{cases}2x-5=0\\2x-6=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=5\\2x=6\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=3\end{cases}}\)
Vậy ...
\(\left(2x-5\right)^2=2x-5\Leftrightarrow\left(2x-5\right)\left(2x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=3\end{cases}}\)