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C1
a) -7x(3x-2)=-21x^2+14x
b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2
C2
a) (x-5)(x+5)
b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0
\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)
Vậy S={-5;2/3}
C3:
a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3
b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)
\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)
4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)
=> (a+b)^2=(a-b)^2+4ab
- 2x – x2 + 2 – x – (3x2 + 6x + 5x +10) = – 4x2 + 2
- 2x – x2 + 2 – x – 3x2 – 6x – 5x – 10 = – 4x2 + 2 –10x = 10 x = – 1
- 2x2 – 6x + x – 3 = 0
(x – 3)(2x + 1) = 0
x = 3 hay x = -1/2
a ) \(\left(x-1\right)\left(x+1\right)-2x^2=0\)
\(\Leftrightarrow x^2-1-2x^2=0\)
\(\Leftrightarrow-x^2-1=0\)
\(\Leftrightarrow-x^2=1\)
\(\Leftrightarrow x^2=-1\) ( Vô lý , \(x^2\ge0\forall x\) )
Vậy ko có g/t x thỏa mãn
b ) \(\left(2x+5\right)\left(x^2-3x+1\right)-x\left(2x^2-1\right)=3\)
\(\Leftrightarrow2x\left(x^2-3x+1\right)+5\left(x^2-3x+1\right)-2x^3+x=3\)
\(\Leftrightarrow2x^3-6x^2+2x+5x^2-15x+5-2x^3+x=3\)
\(\Leftrightarrow\left(2x^3-2x^3\right)-\left(6x^2-5x^2\right)+\left(2x-15x+x\right)+5=3\)
\(\Leftrightarrow-x^2-12x+5=3\)
\(\Leftrightarrow-\left(x^2+12x-5\right)=3\)
\(\Leftrightarrow x^2+12x-5=-3\)
\(\Leftrightarrow x^2+12x+36-41=-3\)
\(\Leftrightarrow\left(x+6\right)^2=-3+41\)
\(\Leftrightarrow\left(x+6\right)^2=38\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=\sqrt{38}\\x+6=-\sqrt{38}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{38}+6\\x=6-\sqrt{38}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\sqrt{38}+6\\x=6-\sqrt{38}\end{matrix}\right.\)
c ) \(\left(x-1\right)2x-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
:D
a, <=> (x-1).(x-6) = 0
<=> x=1 hoặc x=6
b, <=> (x+1).(2x-5) = 0
<=> x=-1 hoặc x=5/2
c, <=> (2x-5).(2x-1) = 0
<=> x=5/2 hoặc x=1/2
d, <=> (x^2-x+1).(x^2+1) = 0
=> pt vô nghiệm vì x^2-x+1 và x^2+1 đều > 0
Tk mk nha
a) x2 - 7x + 6 = 0
<=> x2 - 6x - x + 6 = 0
<=>( x - 6 ) ( x - 1 ) = 0
<=> x - 6 = 0 hoặc x - 1 = 0
1. x - 6 = 0
<=> x = 6
2. x - 1 = 0
<=> x = 1
Vậy ......
b) 2x2 - 3x - 5 = 0
<=> 2x2 + 2x - 5x - 5 = 0
<=> ( x + 1 ) ( 2x - 5 ) = 0
<=> x + 1 = 0 hoặc 2x - 5 = 0
1. x + 1 = 0
<=> x = -1
2. 2x - 5 = 0
<=> x = 2.5
Vậy ............
c) 4x2 - 12x + 5 = 0
<=> 4x2 - 2x - 10x + 5 = 0
<=> 2x ( 2x - 1 ) - 5( 2x - 1 ) = 0
<=> ( 2x - 1 ) ( 2x - 5 ) = 0
<=> 2x - 1 = 0 hoặc 2x - 5 = 0
1. 2x - 1 = 0
<=> x = 0.5
2. 2x - 5 = 0
<=> x = 2.5
Vậy ....................
d) x4 - x3 + 2x2 - x + 1 = 0
a) \(PT\Leftrightarrow x^2-4x+1=3x-5\)
\(\Leftrightarrow x^2-7x+6=0\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
b) \(PT\Leftrightarrow x^2\left(2x-3\right)-\left(2x-3\right)=0\Leftrightarrow\left(x^2-1\right)\left(2x-3\right)=0\Leftrightarrow x\in\left\{\pm1;\frac{3}{2}\right\}\)
a, 3x2 - 8x + 4
= 3x2 - 6x - 2x + 4
= 3x(x - 2) - 2(x - 2)
= (3x - 2)(x - 2)
b, x2 - 4xy + 3y2
= x2 - xy - 3xy + 3y2
= x(x - y) - 3y(x - y)
= (x - 3y)(x - y)
\(a)3x^2-8x+4=3x^2-6x-2x+4=3x\left(x-2\right)-2\left(x-2\right)=\left(3x-2\right)\left(x-2\right)\)
\(b)x^2-4xy+3y^2=x^2-xy-3xy+3y^2=x\left(x-y\right)-3y\left(x-y\right)=\left(x-3y\right)\left(x-y\right)\)
\(c)2x^2+3881x-17505=2x^2+3890x-9x-17505=2x\left(x+1945\right)-9\left(x+1945\right)\)
\(=\left(2x-9\right)\left(x+1945\right)\)
Bài 1 :
1) a2 - 4 + y ( a - 2 )
= ( a + 2 ) ( a - 2 ) + y ( a - 2 )
= ( a - 2 ) ( a + 2 + y )
2) ( x - 2 )2 - 9y2
= ( x - 2 - 3y ) ( x - 2 + 3y )
Bài 2 :
1) 3 ( x + 4 ) - 2x = 5
=> 3x + 12 - 2x = 5
=> x + 12 = 5
=> x = 5 - 12 = - 7
Vậy x = - 7
2) x ( x - 2 ) - x2 - 6 = 0
=> x2 - 2x - x2 - 6 = 0
=> - 2x - 6 = 0
=> 2x = - 6
=> x = \(-\frac{6}{2}=3\)
Vậy x = 3
3 ) x2 - 3x = 0
=> x ( x - 3 ) = 0
=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x\in\left\{0;3\right\}\)
4) 5 - 3 ( x - 6 ) = 4
=> 5 - 3x + 18 = 4
=> 3x = 5 + 18 - 4
=> 3x = 19
=> x = \(\frac{19}{3}\)
Vậy \(x=\frac{19}{3}\)
\(a,x^2-4x+1=0.\)
\(\text{Áp dụng biệt thức }\Delta=b^2-4ac\text{, ta có:}\)(Lớp 9 kì 2 hok)
\(\Delta=-4^2-4.1.1=16-4=12\)
\(\Rightarrow\text{pt có 2 nghiệm }\orbr{\begin{cases}x_1=\frac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x_2=\frac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{cases}}\)
b,bn xem lại đề nếu đúng nói mk 1 tiếng mk làm tiếp cho
Nguyễn Xuân Anh, đề đúng mà