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Bài làm:
Ta có: \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-2^2=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta có : \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)
\(A=4x^2-y^2-2y-1\)
\(=\left(2x\right)^2-\left(y+1\right)^2\)
\(=\left(2x+y+1\right)\left(2x-y-1\right)\)
\(=-197\)
Vậy....
k cho mk nha
x^4-2x^3+3x^2-2x+1
=(x^4-2x^3+x^2)+(x^2-2x+1)
=x^2(x^2-2x+1)+(x^2-2x+1)
=(x^2+1)(x^2-2x+1)
=(x^2+1)(x-1)^2
( a + 2 )3 - a( a - 3 )2
= a3 + 6a2 + 12a + 8 - a( a2 - 6a + 9 )
= a3 + 6a2 + 12a + 8 - a3 + 6a2 - 9a
= 12a2 + 3a + 8
cách của symbolab:
\(\left(a+2\right)^3-a\left(a-3\right)^2\)
\(=a^3+6a^2+12a+8-a\left(a-3\right)^2\)
\(=a^3+6a^2+12a+8-a\left(a^2-6a+9\right)\)
\(=a^3+6a^2+12a+8-a^3+6a^2-9a\)
\(=12a^2+3a+8\)
\(\frac{x+19}{3}+\frac{x+13}{5}=\frac{x+7}{7}+\frac{x+1}{9}\)
\(=>\frac{x+19}{3}+3+\frac{x+13}{5}+3=\frac{x+7}{7}+3+\frac{x+1}{9}+3\)
\(=>\frac{x+28}{3}+\frac{x+28}{5}=\frac{x+28}{7}+\frac{x+28}{9}\)
\(=>\frac{x+28}{3}+\frac{x+28}{5}-\frac{x+28}{7}-\frac{x+28}{9}=0\)
\(=>\left(x+28\right)\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)=0\)
Do :\(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\ne0\)
\(=>x+28=0\)
\(=>x=-28\)
Vậy nghiệm của phương trình trên là : -28
\(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
\(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
a) 3x2 - 5x - 3y2 + 5y
= 3(x2- y2) -5(x-y)
=3(x+y)(x-y) - 5(x-y)
=(x-y)(3x+3y-5)
b) 49 - x2+2xy-y2
= 72 - (x-y)2
=(7-x+y)(7+x-y)
a) \(3x^2-5x-3y^2+5y\)
\(=\left(3x^2-3y^2\right)-\left(5x-5y\right)\)
\(=3\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=3\left[\left(x-y\right).\left(x+y\right)\right]-5\left(x-y\right)\)
\(3\left(x-y\right).\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right).\left[3.\left(x+y\right)-5\right]\)
\(=\left(x-y\right)\left(3x+3y-5\right)\)
b) \(49-x^2+2xy-y^2\)
\(=7^2-x^2+2xy-y^2\)
\(=7^2-\left(x^2-2xy+y^2\right)\)
\(=7-\left(x-y\right)^2\)
\(=\sqrt{7}^2-\left(x-y\right)^2\)
\(=\left[7-\left(x-y\right).-7+\left(x-y\right)\right]\)
\(=\left(7-x+y\right).\left(-7+x-y\right)\)
Ta có : \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x.\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+3x+3=0\end{cases}\Leftrightarrow}x=0\)
x3 + 3x2 + 3x = 0
<=> x( x2 + 3x + 3 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x^2+3x+3=0\left(1\right)\end{cases}}\)
Ta có (1) = x2 + 3x + 3
= ( x2 + 3x + 9/4 ) + 3/4
= ( x + 3/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x
=> (1) vô nghiệm
Vậy phương trình có nghiệm duy nhất là x = 0