(3x-1)/3=(2y-3)/5=(3x-2y-4)/9-x
1, ( x+1/3)^3
2, ( 2x+y^2)^3
3, ( 1/2x^2+1/3y)^3
4, ( 3x^2-2y)^3
5, ( 2/3x^2-1/2y)^3
6, ( 2x+1/2)^3
7, ( x-3)^3
8, ( x+1).(X^2+3x+9)
9, ( x-3).( x^2+3x+9)
10, ( x-2).( x^2+2x+4)
11, ( x+4).( x^2-4x+16)
12, ( x-3y).( x^2+3xy+9y^2)
13, ( x^2-1/3). ( x^4+1/3x^2+1/9)
14, ( 1/3x+2y).( 1/9x^2-2/3xy+4y^2)
Đưa về HĐT
1. (2x+3y)^2
2. (5x-y)^2
3. (2x+y^2)^3
4. (2x-1) (4x^2+2x+1)
5. (5+3x)^3
6.(x^2+2/5y) (x^2-2/5y)
7.(x+1/4)^2
8. (2/3x^2-1/2y)^3
9. (3x^2-2y)^3
10.(x-3y) (x^2+3xy+9y^2)
11. (x^2-3) (x^4+3x^2+9)
12. (x+2y+z) (x+2y-z)
mn giúp em với
a (4-2x)+(5x-3)=(x-2)(x+3)
b 5-3x-(4-3x)=x-7-(x-2)
c 7-1/2y+(1/2y-9/2)=2y+3/4-(1/2+1/2y)
d -3,6-(1,5y+1)=-4y-0,8-(0,4y-2)
e \(\sqrt{3x}-1=x-3\)
f \(2\sqrt{3x}+12=4x-5\sqrt{3}\)
g x+1-3/2x =2+x/2-1/2
Các bạn giỏi Toán vào giúp mình cái:
Tìm giá trị biểu thức: \(P=\frac{3x+2y}{x-2y+4}\) biết\(\frac{2}{x-1}=\frac{3}{2y-3}\)
Thầy giải như sau: \(\frac{2}{x+1}=\frac{3}{2y-3}=\frac{6}{3x+3}=\frac{9}{3x+2y}\)(1)
Mặt khác:\(\frac{2}{x+1}=\frac{3}{2y-3}=\frac{-1}{x+1-2y+3}=\frac{-1}{x-2y+4}\)(2)
Từ (1) và (2) => \(\frac{9}{3x+2y}=\frac{-1}{x-2y+4}\)
=>\(\frac{9}{-1}=\frac{3x+2y}{x-2y+4}=-9\)
Vậy P = -9
Mình thắc mắc chỗ (2) đó các bạn, ở phần \(\frac{-1}{x+1-2y+3}\) đáng lẽ phải là \(\frac{-1}{x+1-2y-3}\) chứ
Giúp mình với
Mình sẽ trình bày rõ hơn ở (2) nha
Ta có:
\(\frac{2}{x+1}=\frac{3}{2y-3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2}{x+1}=\frac{3}{2y-3}\) = \(\frac{2-3}{\left(x+1\right)-\left(2y-3\right)}=\frac{-1}{x+1-2y+3}=\frac{-1}{x-2y+4}\)
(Vì trước ngoặc của 2y - 3 là dấu trừ nên khi phá ngoặc thì nó sẽ trở thành dấu cộng.Đây là quy tắc phá ngoặc mà bạn đã được học ở lớp 6 đó)
Thực hiện phép trừ:
a) 3/x-2-2/x+2
b) 5/2x-3 + 2/2x+3 -2x+5/9-4x^2
c) 2y - 6xy+2y/3x+2y + 2y-9x^2/3x+2y
a: \(=\dfrac{3x+6-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+10}{x^2-4}\)
b: \(=\dfrac{10x+15+4x-6+2x+5}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x+14}{\left(2x-3\right)\left(2x+3\right)}\)
1 :( x^2y + 2y )^3
2:(3x - xy^3)^3
3:(x^2 - 1/3)*(x^4 + 1/3x + 1/9)
1) (x2y + 2y)3
= (x2y)3 + 3.(x2y)2.2y + 3.x2y.(2y)2 + (2y)3
= x6y + 6y.x4y + 12y2 . x2y + 8y3
= ...........
chả bít đúng ko!!! 5476756758568568585787986585685685
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
1)1/x+2y +1/y+2x=3
4/x+2y - 3/y+2x=9
2)3x/x+1 -2/y+4=4
2x/x+1 -5/y+4=9
3x^4 + 3x^2y^2 + 6x^3y - 27x^2
x^4 + x^3 - x^2 + x
2x^5 - 6x^4 - 2a^2x^3 - 6ax^3
x^5 + x^4 + x^3 + x^2 + x + 1
x^3 - 1 + 5x^2 - 5 + 3x - 3
1/4.(a + 1)^2 - 4/9.(a - 2)^2
12a^2b^2 - 3.(a^2b^2)^2
4x^2y^2 - (x^2 + y^2 - a^2)^2
(a + b + c)^2 + (a + b - c)^2 - 4c^2
x^3 - 1 + 5x^2 - 5 + 3x - 3