giải các pt sau:
a, \(\left(x^2+4x+8\right)^2+3x.\left(x^2+4x+8\right)+2x^2=0\) 0
b, \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
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LB
5 tháng 2 2018
b) \(\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)
\(\dfrac{x-2022}{2017}+\dfrac{x-2002}{2020}=\dfrac{x-2022}{2016}+\dfrac{x-2022}{1954}\)
\(\Leftrightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)
\(\Leftrightarrow x-2022=0\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\right)\)
\(\Leftrightarrow x=2022\)
26 tháng 1 2018
a) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)
\(\Rightarrow\left(x^2+4x+8\right)^2+2.\dfrac{3}{2}x\left(x^2+4x+8\right)+\dfrac{9}{4}x^2-\dfrac{1}{4}x^2=0\)
\(\Rightarrow\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\left(\dfrac{1}{2}x\right)^2=0\)
\(\Rightarrow\left(x^2+4x+8+\dfrac{3}{2}x-\dfrac{1}{2}x\right)\left(x^2+4x+8+\dfrac{3}{2}x+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left(x^2+4x+8+x\right)\left(x^2+4x+8+2x\right)=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left(x^2+6x+8\right)=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)=0\)
Vì x2 ≥ 0 với mọi x
⇒ x2 + 5x + 8 ≥ 0 với mọi x
\(\Rightarrow\left(x+2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
26 tháng 1 2018
b) \(\dfrac{x-5}{2017}+\dfrac{x-2}{2020}=\dfrac{x-6}{2016}+\dfrac{x-68}{1954}\)
Trừ 2 vào mỗi vế ta có:
\(\Rightarrow\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)
\(\Rightarrow\dfrac{x-2022}{2017}+\dfrac{x-2022}{2020}-\dfrac{x-2022}{2016}-\dfrac{x-2022}{1954}=0\)
\(\Rightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)
Ta thấy \(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\)
\(\Rightarrow x-2022=0\Rightarrow x=2022\)
Chúc bạn học tốt!
N
31 tháng 1 2019
câu a tự quy đồng cùng mẫu rồi làm thôi :"))
b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
MT
7 tháng 3 2020
a) \(pt\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1-\frac{3x^2+16}{x^2+10}+2=0\)
\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0\)
\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)
\(\Leftrightarrow4-x^2=0\)(do \(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0,\forall x\))
\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
\(KL...\)
7 tháng 3 2020
2x(8x - 1)2(4x - 1) = 9
<=> 512x4 - 256x3 + 40x2 - 2x = 9
<=> 512x4 - 256x3 + 40x2 - 2x - 9 = 0
<=> (2x - 1)(4x + 1)(64x4 - 16x + 9) = 0
vì 64x4 - 16x + 9 khác 0 nên:
<=> 2x - 1 = 0 hoặc 4x + 1 = 0
<=> x = 1/2 hoặc x = -1/4
b) \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)
\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)
\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)
\(\Leftrightarrow\)\(x-2022=0\) (vì 1/2017 + 1/2020 - 1/2016 - 1/1954 \(\ne0\))
\(\Leftrightarrow\)\(x=2022\)
Vậy...
b) \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)
\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)
\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)
\(\Leftrightarrow\)\(x-2022=0\) (vì 1/2017 + 1/2020 - 1/2016 - 1/1954 \(\ne0\))
\(\Leftrightarrow\)\(x=2022\)
Vậy,....