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VD
2
WR
4 tháng 7 2019
\(2x^4-10x^2+17=2\left(x^4-5x^2+\frac{25}{4}\right)+\frac{9}{2}=2\left(x^2-\frac{5}{2}\right)^2+\frac{9}{2}>0\left(vl\right)\)
=> PT vô nghiệm
WR
4 tháng 7 2019
\(x^4-x^3+2x^2-x+1=x^2\left(x^2-x+1\right)+x^2-x+1=\left(x^2-x+1\right)\left(x^2+1\right)=\left(x^2+1\right)\left(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right)>0\forall x\)=> Pt vô nghiệm
BM
0
2 tháng 7 2019
\(a,\frac{7}{x+2}=\frac{3}{x-5}\)
\(\Rightarrow7\left(x-5\right)=3\left(x+2\right)\)
\(\Rightarrow7x-35=3x+6\)
\(\Rightarrow7x-3x=6+35\)
\(\Rightarrow4x=41\)
\(\Rightarrow x=\frac{41}{4}\)
\(b,\frac{2x+5}{2x}-\frac{x}{x+5}=0\)
\(\Rightarrow\frac{2x+5}{2x}=\frac{x}{x+5}\)
\(\Rightarrow\left(2x+5\right)\left(x+5\right)=2x\cdot x\)
\(\Rightarrow2x^2+10x+5x+25=2x^2\)
\(\Rightarrow2x^2+15x+25-2x^2=0\)
\(\Rightarrow15x+25=0\)
\(\Rightarrow15x=-25\)
\(\Rightarrow x=\frac{-5}{3}\)
\(c,\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
\(\Rightarrow\frac{12x+1}{11x-4}=\frac{20x+17}{18}-\frac{10x-4}{9}\)
\(\Rightarrow\frac{12x+1}{11x-4}=\frac{25}{18}\)
\(\Rightarrow\left(12x+1\right)\cdot18=25\cdot\left(11x-4\right)\)
\(\Rightarrow216x+18=275x-100\)
\(\Rightarrow216x-275x=-100-18\)
\(\Rightarrow-59x=-118\)
\(\Rightarrow x=2\)
7 tháng 7 2017
\(3x^4+2x^3-10x^2+2x+3=0\)
\(\Leftrightarrow3x^4-6x^3+3x^2+8x^3-16x^2+8x+3x^2-6x+3=0\)
\(\Leftrightarrow3x^2\left(x^2-2x+1\right)+8x\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(3x^2+8x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(3x^2+8x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(3\left(x+\dfrac{4}{3}\right)^2-\dfrac{7}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3\left(x+\dfrac{4}{3}\right)^2-\dfrac{7}{3}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8\pm\sqrt{28}}{6}\end{matrix}\right.\)
6 tháng 6 2019
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
6 tháng 6 2019
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
21 tháng 7 2016
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x\left(x^2+10\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
21 tháng 7 2016
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x=0\end{cases}}\)(vì \(x^2+10\ge0\) với mọi x)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
3 tháng 8 2020
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
3 tháng 8 2020
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
Đặt t = x2 (t \(\ge\)0)
Khi đó ta có pt: 2t2 - 10t + 17 = 0
<=> 2(t2 - 5t + \(\frac{25}{4}+\frac{9}{4}\)) = 0
<=> 2(t - \(\frac{5}{2}\))2 + \(\frac{9}{2}\)= 0 (VN)
Vậy pt vô nghiệm
\(2x^4-10x^2+17=0\)
\(4x^4-20x^2+68=0\)
\(\left(2x-5\right)^2+33=0\)
\(\left(2x-5\right)^2=-33\)
\(2x-5=\pm\sqrt{33}\)
\(\Rightarrow\orbr{\begin{cases}2x-5=-\sqrt{33}\\2x-5=\sqrt{33}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=5-\sqrt{33}\\2x=5+\sqrt{33}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5-\sqrt{33}}{2}\\x=\frac{5+\sqrt{33}}{2}\end{cases}}}\)