35x – 72 - 12x = 20
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Akai Haruma
Giáo viên
4 tháng 9 2021
Lần sau bạn chú ý ghi đầy đủ yêu cầu đề.
Lời giải:
1. $9x^2+4+12x=(3x)^2+2.3x.2+2^2=(3x+2)^2$
2. Đề sai sai. Bạn xem lại
3.
$(16x^2-4xy+y^2)(4x+y)=(4x+y)[(4x)^2-4x.y+y^2]$
$=(4x)^3+y^3=64x^3+y^3$
4.
$(5x-7)(25x^2+35x+49)=(5x-7)[(5x)^2+5x.7+7^2]$
$=(5x)^3-7^3=125x^3-343$
4 tháng 9 2021
\(9x^2+12x+4=\left(3x+2\right)^2\)
\(\left(16x^2-4xy+y^2\right)\cdot\left(4x+y\right)=64x^3+y^3\)
\(\left(5x-7\right)\left(25x^2+35x+49\right)=125x^3-343\)
14 tháng 7 2019
Sửa đề:
\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+20}{x+6}\)
\(\Leftrightarrow\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)
Quy đồng giải tiếp nhé
LC
1
TV
19 tháng 3 2018
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)ĐKXĐ là \(x\ne-2;x\ne-8;x\ne-4;x\ne-6\)
\(\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+16x+64+8}{x+8}=\dfrac{x^2+8x+16+4}{x+4}+\dfrac{x^2+12x+36+6}{x+6}\)\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\left(\dfrac{2}{x+2}-1\right)+\left(\dfrac{8}{x+8}-1\right)=\left(\dfrac{4}{x+4}-1\right)+\left(\dfrac{6}{x+6}-1\right)\)\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{x}{x+8}-\dfrac{x}{x+4}-\dfrac{x}{x+6}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
Do \(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\ne0\)
=> x=0
Vậy ....
3 tháng 3 2018
\(\Leftrightarrow1+\dfrac{2}{x+2}+1+\dfrac{8}{x+8}=1+\dfrac{4}{x+4}+1+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{4}{x+8}=\dfrac{2}{x+4}+\dfrac{3}{x+6}\)
\(\Leftrightarrow\dfrac{4}{x+8}-\dfrac{3}{x+6}=\dfrac{2}{x+4}-\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{4x+24-3\left(x+8\right)}{\left(x+8\right)\left(x+6\right)}=\dfrac{2x+4-\left(x+4\right)}{\left(x+4\right)\left(x+2\right)}\)
\(\dfrac{x}{\left(x+8\right)\left(x+6\right)}=\dfrac{x}{\left(x+4\right)\left(x+2\right)}\)
x=0 là nghiệm
x khác 0
\(\left\{{}\begin{matrix}x\ne\left\{-8;-6;-4;-2\right\}\\\left(x+4\right)\left(x+2\right)=\left(x+8\right)\left(x+6\right)\end{matrix}\right.\)<=>x^2 +6x+8 =x^2 +14x+48
-40 =8x=> x =-5 nhận
x={-5;0}
P
2 tháng 1 2018
=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)
=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0
=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)
Voi \(\frac{1}{x+2}-....\)=0 ta co
Dat x+5=t
=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0
=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)
=>t=0
=>x=-5
Vay phuong trinh co nghiem x=0;-5
11 tháng 3 2018
=> \(\frac{(x+2)^2+2}{x+2}+\frac{(x+8)^2+8}{x+8}=\frac{(x+4)+4}{x+4}+\frac{(x+6)^2+6}{x+6}\)
=> 2x + 10 + \(\frac{2}{x+2}+\frac{8}{x+8}=2x+10+\frac{4}{x+4}+\frac{6}{x+6}\)
=>-x \((\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}-\frac{1}{x+8})=0\)
\(x=0\)
\(=>\orbr{\frac{1}{x+2}}-.....+\frac{1}{x+8}=0\)
Với \(\frac{1}{x+2}-...=0\). Ta có :
Đặt x + 5 = t
=> \(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}=0\)
\(=>2t(\frac{1}{t^2-1}+\frac{1}{t^2-9})=0\)
=> t = 0
=> x = -5
Vậy phương trình có nghiệm x= 0 ; - 5
15 tháng 4 2018
\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\left(đkxđ:x\ne-2;-8;-4;-6\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{2}{x+2}-1+\dfrac{8}{x+8}-1=\dfrac{4}{x+4}-1+\dfrac{6}{x+6}-1\)
\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)
\(\Leftrightarrow\left(-x\right)\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
`35x-72-12x=20`
`35x-12x=20+72`
`(35-12)x=92`
`23x=92`
`x=92:23=4`
35x - 72 - 12x = 20
x(35-12) - 72 = 20
x23 - 72 = 20
x23 = 20 + 72
x23 = 92
x = 92 : 23
x = 4