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PL
6 tháng 3 2020
a) x4 - 5x2 + 4 = 0 (*)
đặt x2 = m (\(m\ge0\))
(*) <=> m2 - 5m + 4 = 0
m2 - 4m - m + 4 = 0
m(m - 4) - (m - 4) = 0
(m - 4)(m - 1) = 0
vậy m - 4 = 0 hoặc m - 1 = 0
hay m = 4 hoặc m = 1
m = 4 => x2 = 4 => \(x=\pm2\)
m = 1 => x2 = 1 => \(x=\pm1\)
TT
6 tháng 3 2020
d) \(x\left(x+1\right)\left(x-1\right)\left(x-2\right)=24\)
\(\Leftrightarrow\left[x\left(x-1\right)\right]\left[\left(x+1\right)\left(x-2\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right)\left(x^2-x-2\right)-24=0\)
\(\Leftrightarrow\left(x^2-x\right)^2-2\left(x^2-x\right)+1-25=0\)
\(\Leftrightarrow\left(x^2-x+1\right)^2-25=0\)
\(\Leftrightarrow\left(x^2-x+6\right)\left(x^2-x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-x+6=0\left(1\right)\\x^2-x-4=0\left(2\right)\end{cases}}\)
+) Pt (1) \(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=-\frac{23}{4}\) ( vô nghiệm )
+) Pt (2) \(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{17}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{17}}{4}+\frac{1}{2}\\x=-\frac{\sqrt{17}}{4}+\frac{1}{2}\end{cases}}\) ( thỏa mãn )
Vậy pt đã cho có nghiệm \(S=\left\{\pm\frac{\sqrt{17}}{4}+\frac{1}{2}\right\}\)
TN
2 tháng 3 2016
f(x)g(x)=0<=>f(x)=0 hoặc g(x)=0
<=>(x2-5x)2+10(x2-5x)+24=(x-4)(x-3)(x-2)(x-1)
TH1:x-4=0
=>x=4
TH2:x-3=0
=>x=3
TH3:x-2=0
=>x=2
TH4:x-1=0
=>x=1
vậy giá trị nguyên của x lần lượt là {1;2;3;4}
8 tháng 1 2021
a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy: S={1;2;3;4}
b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)
c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\forall x\)
nên (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: S={2;-1}
d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
NH
Giải các phương trình:
\(a,\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(b,x^4-30x^2+31x-30=0\)
2
3 tháng 5 2019
a, Đặt \(x^2-5x=a\)
\(\Rightarrow\)\(a^2+10a+24=0\)
\(\Rightarrow a^2+4a+6a+24=0\)
\(\Rightarrow\left(a+4\right)\left(a+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+4=0\\a+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2-5x+4=0\left(1\right)\\x^2-5x+6=0\left(2\right)\end{cases}}}\)
Giải pt (1) ta có : \(x^2-5x+4=0\)
\(\Rightarrow x^2-4x-x+4=0\)
\(\Rightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Giải pt (2) ta có : \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy \(S=\left\{1;2;3;4\right\}\)
3 tháng 5 2019
\(x^4-30x^2+31x-30=0\)
\(\Rightarrow x^4-30x^2+x+30x-30=0\)
\(\Rightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
Mà \(x^2-x+1>0\)với \(\forall\)\(x\)
\(\Rightarrow x^2+x-30=0\)
\(\Rightarrow x^2-5x+6x-30=0\)
\(\Rightarrow x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Vậy \(S=\left\{5;-6\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2020
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2020
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
EC
1
MS
3 tháng 1 2018
câu a bạn sai đề nha
b)
\(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)
\(x^4+x^2+1+2x^3+2x^2+2x=3x^4+3x^2+3\)
\(2\left(x^3+x^2+x\right)=2\left(x^4+x^2+1\right)\)
\(x^4-x^3+1-x=0\)
\(x^3\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^3-1\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\x^3-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{1\right\}\)
NT
3
HT
22 tháng 8 2016
a) \(x^3\)+\(x^2\)=36
\(\Leftrightarrow\)\(x^3\)+\(x^2\)\(-36=0\)
\(\Leftrightarrow\)\(x^3\)\(-3x^2\)\(+4x^2\)\(-12x\)\(+12x-36=0\)
\(\Leftrightarrow\)\(x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2+4x+12\right)=0\)
Suy ra: \(x-3=0\) hoặc \(x^2+4x+12=0\)
- \(x-3=0\) \(\Leftrightarrow\) \(x=3\)
- \(x^2+4x+12=0\) (phương trình vô nghiệm)
Vậy \(x=3\)