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a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)2
làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
a) Ta có:
VT = (x - y)² + 4xy
= x² - 2xy + y² + 4xy
= x² + 2xy + y²
= (x + y)²
= VP
b) Ta có:
(x + y)² = (x - y)² + 4xy
= 5² + 4.3
= 25 + 12
= 37
a : 7 dư 3 cm a2 : 7 dư 2
Ta có: a = 7k + 3
⇔ a2 = (7k + 3)2
⇔ a2 = 49k2 + 42k + 9
⇔ a2 = 7.(7k2 + 6k + 1) + 2
7 ⋮ 7 ⇔ 7.(7k2 + 6k + 1) ⋮ 7
⇔ a2 = 7.(7k2 + 6k + 1) + 2 : 7 dư 2 (đpcm)
Cách 2 sử dụng đồng dư thức:
a \(\equiv\) 3 (mod 7) ⇔ a2 \(\equiv\) 32 (mod 7) 32 : 7 dư 2 ⇔ a2 : 7 dư 2 (đpcm)
a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
a, F(\(x\)) = a\(x^2\) + b\(x\) + c (a; b; c \(\in\) Q và a \(\ne\) 0)
Vì F(\(x\)) có nghiệm là \(\sqrt{2}\) ta có F(\(\sqrt{2}\)) = 0
⇔ a.(\(\sqrt{2}\))2 + b.(\(\sqrt{2}\)) + c = 0
2a + \(\sqrt{2}\)b + c = 0 ⇒ c = - (2a + \(\sqrt{2}\)b) (1)
a\(x^2\) + b\(x\) + c = 0
a(\(x^2\) + 2. \(\dfrac{b}{2a}\)\(x\) + \(\dfrac{b^2}{4a^2}\)) - \(\dfrac{b^2-4ac}{4a}\) = 0
a.(\(x\) + \(\dfrac{b}{2a}\))2 = \(\dfrac{b^2-4ac}{4a}\)
(\(x\) + \(\dfrac{b}{2a}\) )2 = \(\dfrac{b^2-4ac}{4a^2}\)
\(\left[{}\begin{matrix}x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\end{matrix}\right.\)
Thay (1) vào \(x\) = \(\dfrac{-b-\sqrt{b^2-4ac}}{2a}\) ta có
\(x\) = \(\dfrac{-b-\sqrt{b^2-4a\left(2a+\sqrt{2}b\right)}}{2a}\)
a) \(f\left(x\right)=ax^2+bx+c=0\)
\(\Rightarrow f\left(x_1=\sqrt[]{2}\right)=2a+b\sqrt[]{2}+c=0\left(1\right)\)
\(S=x_1+x_2=-\dfrac{b}{a}\Rightarrow x_2=-\dfrac{b}{a}-x_1=-\dfrac{b}{a}-\sqrt[]{2}\)
\(P=x_1.x_2=\dfrac{c}{a}\Rightarrow x_2=\dfrac{c}{a.x_1}=\dfrac{c}{a.\sqrt[]{2}}\)
Vậy nghiệm còn lại là \(-\dfrac{b}{a}-\sqrt[]{2}\) hay \(\dfrac{c}{a.\sqrt[]{2}}\left(a,b,c\in Q;a\ne0\right)\)
b) \(P\left(x\right)=x^2-px+q\)
\(S=x_1+x_2=p;P=x_1.x_2=q\)
Để P(x) có nghiệm \(x_1;x_2\) đều là số nguyên
\(\Rightarrow S=p;P=q\) đều là số nguyên
mà \(p,q\) là số nguyên tố
\(\Rightarrow p;q⋮1\)
\(\Rightarrow\left(p;q\right)\in\left\{-1;1\right\}\Rightarrow p=\pm1;q=\pm1\)
Ta thay \(p=\pm1;q=\pm1\) vào \(P\left(x\right)=x^2-px+p=0\) ta được \(\Delta=5;\Delta=-4< 0\) \(\Rightarrow p,q\) không thỏa nghiệm đa thức nguyên
\(\Rightarrow\left(p;q\right)\in\varnothing\)
Hình của em đâu, phần tô màu là phần nào thì mới chứng minh chính xác được em nhé
