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x + x2 = 0
=> x(1 + x) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
vậy_
mk biến đổi về pt tích, sau đó bạn tính nốt nhé:
b) \(x+1-\left(x+1\right)^2=0\)
<=> \(\left(x+1\right)\left(1-x-1\right)=0\)
<=> \(-x\left(x+1\right)=0\)
c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
<=> \(3\left(4y-9\right)\left(5y-1\right)=0\)
d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
<=> \(\left(25z+7\right)\left(8-27z\right)=0\)
a) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
\(\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}15y-3=0\\4y-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{5}\\y=\frac{9}{4}\end{matrix}\right.\)
Vây \(y\in\left\{\frac{1}{5};\frac{9}{4}\right\}\)
b) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
\(\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}z=\frac{8}{27}\\z=-\frac{7}{25}\end{matrix}\right.\)
Vậy \(z\in\left\{\frac{8}{27};-\frac{7}{25}\right\}\)
c) \(13y\left(y-8\right)-2y+16=0\)
\(\Leftrightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)
\(\Leftrightarrow\left(13y-2\right)\left(y-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{2}{13}\\y=8\end{matrix}\right.\)
Vậy \(y\in\left\{\frac{2}{13};8\right\}\)
d) \(-10y\left(y+2\right)-y-2=0\)
\(\Leftrightarrow\left(-10y-1\right)\left(y+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\frac{1}{10}\end{matrix}\right.\)
Vậy \(y\in\left\{-2;-\frac{1}{10}\right\}\)
e) \(x\left(x+19\right)^2-\left(x+19\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+19\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{1;-19\right\}\)
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
Bài 3:
\(a,x^2-81=0\)
\(\Rightarrow x^2-9^2=0\)
\(\Rightarrow\left(x-9\right)\left(x+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
\(b,x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)
\(\Rightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
bài 1: rút gọn biểu thức:
B = (x−2y)2- (x+2y)2+ (4y + 1) ( 1 - 4y)
= x2 - 4xy+ 4y2 - x2 +4xy+4y2+4y- 16y2 +1-4y
=2x2- 8y2+1
E = (2x−3)2 - (3x+1)2 - 5 (x-2) (x+2)
=4x2- 12x+ 9- 9x2+ 6x+ 1- 5x2+20
= - 10x2- 6x+ 30
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x\right)^2-5^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(-2\right).\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
a,\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(4x^2-25\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-5-2x-7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(-12\right)=0\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
\(b,2x^3+3x^2+2x+3=0\)
\(\Rightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Rightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-3\\x^2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
\(c,x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right).x^3=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)
\(d,x^2\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
Lời giải:
a)
Ta có: \(x^2+10x+30=x^2+2.x.5+5^2+5=(x+5)^2+5\)
Vì $(x+5)^2\geq 0, \forall x\Rightarrow x^2+10x+30=(x+5)^2+5\geq 5>0$ (đpcm)
b)
\(4x-x^2-7=-(x^2-4x+7)=-(x^2+4x+4+3)=-[(x-2)^2+3]\)
Vì $(x-2)^2\geq 0, \forall x\Rightarrow (x-2)^2+3\geq 3>0$
$\Rightarrow 4x-x^2-7=-[(x-2)^2+3]< 0$ (đpcm)
c)
\(x^2+4y^2-2x-4y+2=(x^2-2x+1)+(4y^2-4y+1)\)
\(=(x-1)^2+(2y-1)^2\)
Vì $(x-1)^2\geq 0; (2y-1)^2\geq 0, \forall x,y$
$\Rightarrow x^2+4y^2-2x-4y+2=(x-1)^2+(2y-1)^2\geq 0$ (đpcm)
Câu 1: D. \(\frac{1}{2}-4x=0\)
Câu 2: C. 2x - 1 = x
Câu 3: D. S = {-9}
# Chúc bạn học tốt #
a) \(x+x^2=0\Leftrightarrow x\left(1+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
b) \(x+1-\left(x+1\right)^2=0\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{15}=\dfrac{1}{5}\\x=\dfrac{9}{4}\end{matrix}\right.\)
d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}z=\dfrac{8}{27}\\z=\dfrac{-7}{25}\end{matrix}\right.\)