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Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
Bài 1:
a) \(x\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
b) \(\left(3-x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-0\\x^2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\sqrt{-1}\end{matrix}\right.\)
Vì một số không âm mới có căn bậc hai \(\Rightarrow x=3\)
c) \(\left(x-1\right)^2=4\)
\(\Leftrightarrow\left(x-1\right)^2=2^2=\left(-2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+1\\x=\left(-2\right)+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
d) \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
\(\Leftrightarrow2x=2+1\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\dfrac{3}{2}\) (không thỏa mãn vì \(x\in Z\))
Vậy \(x\in\varnothing\)
e) \(\left(x+3\right)^2=16\)
\(\Leftrightarrow\left(x+3\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4-3\\x=\left(-4\right)-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
a: \(A=\left\{0;1;2;3;4;5\right\}\)
b: \(B=\left\{2;3;4;5\right\}\)
c: \(C=\left\{0;1;-1;2;-2;3;-3\right\}\)
6:
k: =>x^2-9<x^2+2x+3
=>2x+3>-9
=>2x>-12
=>x>-6
1:
h: =>x(x-1)=0
=>x=0; x=1
i: =>x(x-3)=0
=>x=0; x=3
Lời giải:
a. $(x^2-9)(5x+15)=0$
$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$
$\Rightarrow x^2=9=3^2=(-3)^2$
$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$
$\Rightarrow x=-3$
b.
$x^2-8x=0$
$\Rightarrow x(x-8)=0$
$\Rightarrow x=0$ hoặc $x-8=0$
$\Rightarrow x=0$ hoặc $x=8$
c.
$5+12(x-1)^2=53$
$12(x-1)^2=53-5=48$
$(x-1)^2=48:12=4=2^2=(-2)^2$
$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$
d.
$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$
$\Rightarrow x=11$ hoặc $x=-1$
e.
$(3x-5)^3=64=4^3$
$\Rightarrow 3x-5=4$
$\Rightarrow 3x=9$
$\Rightarrow x=3$
f.
$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$
$2^{4x}(1+8)=144$
$2^{4x}.9=144$
$2^{4x}=144:9=16=2^4$
$\Rightarrow 4x=4\Rightarrow x=1$
c/ 
d/ 
)(x-
) = 0 g/ (3x-1)(2x-3)(x+5) = 0 
h)
m)
= 8 – x k) 
= – 4x +7
q) 
Bài 1.
a ) \(x\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right..\)
Vậy .....
b ) \(\left(3-x\right)\left(x^2+1\right)=0\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy ........
c ) giống câu b
d ) \(\left(2x-1\right)^3=8\Leftrightarrow2x-1=2\Leftrightarrow\Leftrightarrow x=\dfrac{3}{2}\)
e ) \(\left(x+3\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.
Vậy mà ko bt để khỏi phải gửi câu hỏi nữa
