\(^{x^2-5x+4=0}\)
\(x^2+15x+56=0\)
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26 tháng 3 2020
b, - ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Ta có : \(\frac{5x}{x^2-4}-\frac{4}{x+2}=\frac{5}{x-2}\)
=> \(\frac{5x}{x^2-4}-\frac{4\left(x-2\right)}{x^2-4}=\frac{5\left(x+2\right)}{x^2-4}\)
=> \(5x-4\left(x-2\right)=5\left(x+2\right)\)
=> \(5x-4x+8=5x+10\)
=> \(5x-4x-5x=10-8\)
=> \(-4x=2\)
=> \(x=-\frac{1}{2}\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-\frac{1}{2}\right\}\)
c, Ta có : \(x^4-15x^2+56=0\)
=> \(\left(x^2\right)^2-\frac{2.x^2.15}{2}+\frac{225}{4}-\frac{1}{4}=0\)
=> \(\left(x^2-\frac{15}{2}\right)^2=\frac{1}{4}\)
=> \(\left[{}\begin{matrix}x^2-\frac{15}{2}=\sqrt{\frac{1}{4}}\\x^2-\frac{15}{2}=-\sqrt{\frac{1}{4}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x^2=\sqrt{\frac{1}{4}}+\frac{15}{2}=8\\x^2=-\sqrt{\frac{1}{4}}+\frac{15}{2}=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\\x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{\sqrt{8};-\sqrt{8};\sqrt{7};-\sqrt{7}\right\}\)
26 tháng 3 2020
a)
\(\frac{x-5x-1}{6}=\frac{8-3x}{4}\)
\(\Leftrightarrow\frac{4x-20x-4}{24}=\frac{48-18x}{24}\)
\(\Leftrightarrow\frac{-16x-4}{24}=\frac{48-18x}{24}\)
\(\Leftrightarrow\frac{-16x-4-48+18x}{24}=0\)
\(\Leftrightarrow\frac{2x-52}{24}=0\)
\(\Rightarrow2x-52=0\)
\(x=\frac{52}{2}=26\)
US
1
7 tháng 2 2020
Ta có:
(x2 - 3x + 2)(x2 + 15x + 56) + 8 = 0
\(\Leftrightarrow\) [(x - 2)(x - 1)][(x + 7)(x + 8)] + 8 = 0
\(\Leftrightarrow\) [(x - 2)(x + 8)][(x - 1)(x + 7)] + 8 = 0
\(\Leftrightarrow\) (x2 + 6x - 16)(x2 + 6x - 7) + 8 = 0 (*)
Đặt x2 + 6x - 16 = a \(\Leftrightarrow\) a = (x + 3)2 - 25 \(\ge\) -25
Phương trình (*) trở thành:
a(a + 9) + 8 = 0
\(\Leftrightarrow\) 4a2 + 36a + 32 = 0
\(\Leftrightarrow\) (2a + 9)2 = 49
\(\Leftrightarrow\) \(\left[{}\begin{matrix}a=-1\left(TMĐK\right)\\a=-8\left(TMĐK\right)\end{matrix}\right.\)
+) Nếu a = -1 thì (x + 3)2 - 25 = -1
\(\Leftrightarrow\) x = \(\pm\sqrt{24}-3\)
+) Nếu a = -8 thì (x + 3)2 - 25 = -8
\(\Leftrightarrow\) x = \(\pm\sqrt{17}-3\)
Vậy...
HN
2
18 tháng 5 2021
2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)
\(\Leftrightarrow x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0\)
\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-5x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)
18 tháng 5 2021
1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)
\(\Leftrightarrow x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x+3\right)+\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x^2+x+3\right)=0\)
mà \(x^2+x+3>0\forall x\)
nên (x+1)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy: S={-1;-3}
3D
Giải phương trình
6(2x-5)-6x2+15x=0
\(\dfrac{x+2}{x}\)=\(\dfrac{x^2+5x+4}{x^2+2x}\)+\(\dfrac{x}{x+2}\)
1
7 tháng 4 2022
a: =>6(2x-5)-3x(2x-5)=0
=>(2x-5)(6-3x)=0
=>x=5/2 hoặc x=2
b: \(\Leftrightarrow x^2+5x+4+x^2=x^2+4x+4\)
=>x2+x=0
=>x(x+1)=0
=>x=0(loại) hoặc x=-1(nhận)
15 tháng 5 2021
`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`
15 tháng 5 2021
2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`
YN
4 tháng 10 2021
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
\(x^2-5x+4=0\)
\(\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)
\(x^2+15x+56=0\)
\(\Leftrightarrow x^2+7x+8x+56=0\)
\(\Leftrightarrow x\left(x+7\right)+8\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+7=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\x=-8\end{cases}}}\)
=.= hk tốt!!
1) x2 - 5x + 4 = 0
<=> (x - 1)(x - 4) = 0
<=> x - 1 = 0 hoặc x - 4 = 0
x = 0 + 1 x = 0 + 4
x = 1 x = 4
=> x = 1 hoặc x = 4
2) x2 + 15x + 56 = 0
<=> (x + 7)(x + 8) = 0
<=> x + 7 = 0 hoặc x + 8 = 0
x = 0 - 7 x = 0 - 8
x = -7 x = -8
=> x = -7 hoặc x = -8