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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
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
1.
a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)
b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)
2.
a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)
= (9x-3y)(3x-y)
= 3(3x-y)(3x-y)
= 3(3x-y)^2
b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)
= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)
= \(\left(x^2-9\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)^2\)
Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)
\(=-2x^5-10x^4+x^3\)
b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)
\(=3x^2-5x+2\)
2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)
\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)
\(=\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(x-y\right)\)
b ) \(x^3-3x^2-9x+27\)
\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)
Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
Ta có : 6x(3x + 5) - 2x(9x - 2) = 17
<=> 18x2 + 30x - 19x2 + 4x = 17
<=> 34x2 = 17
=> x2 = 17 : 34
=> \(x^2=\frac{1}{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
\(a,6x\cdot\left(3x+5\right)-2x\cdot\left(9x-2\right)=17\)
\(\Leftrightarrow18x^2+30x-18x^2+4x=17\)
\(\Leftrightarrow34x=17\)
\(\Leftrightarrow x=\frac{1}{2}\)
\(b,2x\cdot\left(3x-1\right)-3x\cdot\left(2x+11\right)-70=0\)
\(\Leftrightarrow6x^2-2x-6x^2-33x-70=0\)
\(\Leftrightarrow-35x=70\)
\(\Leftrightarrow x=-2\)
Đầu bài ý c là j vậy ... mình k thấy zõ
Chúc bạn học giỏi
Kết bạn với mình nha
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) \(PT\Leftrightarrow x^2\left(2x-3\right)-\left(2x-3\right)=0\Leftrightarrow\left(x^2-1\right)\left(2x-3\right)=0\Leftrightarrow x\in\left\{\pm1;\frac{3}{2}\right\}\)