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2.a) \(8x^2-4x=0\Rightarrow4x\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(5x\left(x-3\right)+7\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(5x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\5x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1.4\end{matrix}\right.\)
c) \(2x^2=x\Rightarrow2x^2-x=0\Rightarrow x\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0.5\end{matrix}\right.\)
d) \(x^3=x^5\Rightarrow x^3-x^5=0\Rightarrow x^3\left(1-x^2\right)=0\\ \Rightarrow x^3\left(1-x\right)\left(1+x\right)=0\Rightarrow\left[{}\begin{matrix}x^3=0\\1-x=0\\1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+2x\right)=0\Rightarrow\left(x+1\right)x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
g. \(x\left(2x-3\right)-2\left(3-2x\right)=0\)
\(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1.5\\x=-2\end{matrix}\right.\)
a: \(=4a^3b^3c\left(10xc+3bc-5ab^2x\right)\)
b: \(=\left(b-2c\right)\left(a-b\right)+\left(a+b\right)\left(b-2c\right)\)
=(b-2c)(2a)
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
1. a) 7x2 - 5x - 2 = 7x2 - 7x + 2x - 2 = 7x(x - 1) + 2(x - 1) = (x - 1).(7x + 2)
2. 5(2x - 1)2 - 3(2x - 1) = 0
<=> (2x - 1).[5(2x - 1) - 3] = 0
<=> (2x - 1).(10x - 8) = 0
<=> (2x - 1) = 0 hoặc (10x - 8) = 0
<=> x = 1/2 hoặc x = 4/5
3. x2 - 4x + 7 = (x2 - 4x + 4) + 3 = (x - 2)2 + 3
Do: (x - 2)2 > hoặc = 0 (với mọi x)
Nên (x - 2)2 + 3 > hoặc = 3 (với mọi x)
Hay (x - 2)2 + 3 > 0 (với mọi x) => đpcm
1.
a.\(\Leftrightarrow7x-5x=3+12\)
\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)
b.\(\Leftrightarrow6x-10-7x-7=2\)
\(\Leftrightarrow x=-19\)
c.\(\Leftrightarrow1-3x=4x-3\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)
d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)
\(\Leftrightarrow-2=12\left(voli\right)\)