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11 tháng 7 2017
giải
5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0
5x-(4x+8-2x^2-4x+x^3+2x^2)+x(x^2-1)=0
5x-4x-8+2x^2+4x-x^3-2x^2+x^3-1x=0
(5x-4x+4x-1x)+(-8)+(2x^2-2x^2)+(-x^3+x^3)=0
4x+(-8)=0
4x=0+8
4x=8
x=8:4
x=2
11 tháng 7 2017
D)(4x+1)(16x^2-4x+1)-16x(4x^2-5)=17
64x^3-16x^2+4x+16x^2-4x+1-64x^3+80x=17
80x+1=17
80x=17-1
80x=16
x=1/5
LA
1
11 tháng 8 2023
a) \(4x^2+16x+3=0\)
\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)
b) \(7x^2+16x+2=1+3x^2\)
\(4x^2+16x+1=0\)
\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)
c) \(4x^2+20x+4=0\)
\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)
27 tháng 3 2020
a) Ta có: \(x^2-9x+20=0\)
\(\Leftrightarrow x^2-5x-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
Vậy: x∈{4;5}
b) Ta có: \(x^3-4x^2+5x=0\)
\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)
Ta có: \(x^2-4x+5\)
\(=x^2-4x+4+1=\left(x-2\right)^2+1\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)
hay \(x^2-4x+5>0\forall x\)(2)
Từ (1) và (2) suy ra x=0
Vậy: x=0
c) Sửa đề: \(x^2-2x-15=0\)
Ta có: \(x^2-2x-15=0\)
\(\Leftrightarrow x^2+3x-5x-15=0\)
\(\Leftrightarrow x\left(x+3\right)-5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
Vậy: x∈{-3;5}
d) Ta có: \(\left(x^2-1\right)^2=4x+1\)
\(\Leftrightarrow x^4-2x^2+1-4x-1=0\)
\(\Leftrightarrow x^4-2x^2-4x=0\)
\(\Leftrightarrow x\left(x^3-2x-4\right)=0\)
\(\Leftrightarrow x\left(x^3+2x^2+2x-2x^2-4x-4\right)=0\)
\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)
\(\Leftrightarrow x\cdot\left(x^2+2x+2\right)\cdot\left(x-2\right)=0\)(3)
Ta có: \(x^2+2x+2\)
\(=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)
hay \(x^2+2x+2>0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: x∈{0;2}
TN
21 tháng 6 2017
\(p=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2010\)\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2010=x^3+1+x-x^3+1+2010=x+2012\)Với \(x=-2010\Rightarrow p=-2010+2012=2\)
\(q=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)=64x^3-80x-64x^3-1=-80x-1\)Với \(x=\dfrac{1}{5}\Rightarrow q=-80.\dfrac{1}{5}-1=-17\)
NC
Tìm x biết:
4x2 - 6x = 0
b) 4x2 + 4x = -1
c) 5x2 + x = 0
d) x3 - 5x = 4x2
3x(x-2) = x-2
x3 - 16x = 0
1
LT
6 tháng 9 2020
Tìm x biết:
4x2 - 6x = 0
\(\Leftrightarrow2x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;\frac{3}{2}\right\}\)
b) 4x2 + 4x = -1
\(\Leftrightarrow4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
c) 5x2 + x = 0
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{5}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{1}{5}\right\}\)
d) x3 - 5x = 4x2
\(\Leftrightarrow x^3-4x^2-5x=0\)
\(\Leftrightarrow x^3+x^2-5x^2-5x=0\)
\(\Leftrightarrow x^2\left(x+1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-5x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=5\end{matrix}\right.\)
Vậy x ={0; - 1; 5}
3x(x-2) = x-2
\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=\left\{2;\frac{1}{3}\right\}\)
x3 - 16x = 0
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy x = {0; 4; -4}
15 tháng 2 2020
\(a.\left(3-x\right)^2-12+4x=0\)
\(\Rightarrow\left(3-x\right)^2-4.\left(3-x\right)=0\)
\(\Rightarrow\left(3-x\right)\left(-x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\-x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(b.\left(4x-5\right)^2-2.\left(16x^2-25\right)=0\)
\(\Rightarrow\left(4x-5\right)^2-2.\left(4x+5\right).\left(4x-5\right)=0\)
\(\Rightarrow\left(4x-5\right)\left(4x-5-8x-10\right)=0\)
\(\Rightarrow\left(4x-5\right)\left(-4x-15\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x-5=0\\-4x-15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}\)