Mấy bn cho mik hỏi câu này vs
x^3+4x^2+4x+3
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7(x - 3) - x(3 - x)
= (x - 3)(7 + x)
chỉ bt có v mà k bt có đúng k
1 ) 7 ( x - 3 ) - x ( 3 - x )
= 7 ( x - 3 ) + x ( x - 3 )
= ( x - 3 ) ( 7 + x )
2 ) 4x2 - 6x + 3 - 2x
= 4x2 - 2x - 6x + 3
= 2x ( 2x - 1 ) - 3 ( 2x - 1 )
= ( 2x - 1 ) ( 2x - 3 )
3 ) ( 4 - x ) - 4x + x2
= ( 4 - x ) - x ( 4 - x )
= ( 4 - x ) ( 1 - x )
4 ) x2 - 2xy + y2
= ( x - y )2
\(\sqrt{4x^2-4x+9}=3\\ \Rightarrow4x^2-4x+9=9\\ \Rightarrow4x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Ta có: \(\sqrt{4x^2-4x+9}=3\)
\(\Leftrightarrow4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a) \(\left(3x-2^4\right).7^3=2.7^4\)\(\Leftrightarrow3x-2^4=2.7^4:7^3\)
\(\Leftrightarrow3x-16=2.7\)\(\Leftrightarrow3x-16=14\)\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\)
b) \(3x+4x=\left|-75\right|+23\)\(\Leftrightarrow7x=75+23\)
\(\Leftrightarrow7x=98\)\(\Leftrightarrow x=14\)
Vậy \(x=14\)
a) \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
=> \(3x\cdot7^3-2^4\cdot7^3=2\cdot7\cdot7^3\)
=> \(3x\cdot7^3=14\cdot7^3+16\cdot7^3\)
=> \(3x\cdot7^3=\left(14+16\right)\cdot7^3\)
=> \(3x\cdot7^3=30\cdot7^3\)
=> \(3x=30\)(bỏ hai vế 73)
=> \(x=10\)
Vậy x = 10
b) \(3x+4x=\left|-75\right|+23\)
=> \(7x=75+23\)
=> \(7x=98\)
=> \(x=14\)
Vậy x = 14
3x(12x-4)-(4x-3)(9x+4) = 9
36x2 -12x-(36x2 -16x-27x-12) = 9
36x2 -12x-36x2 -16x+27x+12 = 9
-x = 9-12
-x = -3
x= -3 : -1
x= 3
vậy x= 3
a) \(x^3-7x-6\)
\(=\left(x^3+2x^2\right)-\left(2x^2+4x\right)-\left(3x+6\right)\)
\(=\left(x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x+1\right)\)
b)\(x^3-19x-30\)
\(=\left(x^3-5x^2\right)+\left(5x^2-25x\right)+\left(6x-30\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
c) \(a^3-6a^2+11a-6\)
\(=\left(a^3-a^2\right)-\left(5a^2-5a\right)+\left(6a-6\right)\)
\(=\left(a-1\right)\left(a^2-5a+6\right)\)
\(=\left(a-1\right)\left(a-2\right)\left(a-3\right)\)
\(m\left(x\right)+h\left(x\right)=g\left(x\right)-5\)
\(\Leftrightarrow m\left(x\right)=g\left(x\right)-h\left(x\right)-5\)
\(\Leftrightarrow m\left(x\right)=4x^2+3x+1-3x^2+2x+3-5\)
\(\Leftrightarrow m\left(x\right)=x^2+5x-1\)
\(2x-1-x^2=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\\ \left(1-3\right)^3-1=\left(-2\right)^3-1=-2-1=-3\\ \left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\\ \left(x+2\right)^3+1=\left(x+2+1\right)\left[\left(x+2\right)^2+\left(x+2\right)+1\right]\\ =\left(x+3\right)\left(x^2+4x+4+x+2+1\right)\\ =\left(x+3\right)\left(x^2+5x+7\right)\)
\(x^3+4x^2+4x+3\)
\(=\left(x^3+3x^2\right)+\left(x^2+3x\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+x+1\right)\)
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