3x+4y=36
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a)x3-6x2+9x=x(x2-6x+9)=x(x-3)2
b)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
c)x2-x+xy-y=x(x-1)+y(x-1)=(x-1)(x+y)
d)3x2-6xy-75+3y2=3[(x2-2xy+y2)-25]=3[(x-y)2-52]=3(x-y-5)(x-y+5)
e)2x2-5x-7=(2x2+2x)-(7x+7)=2x(x+1)-7(x+1)=(x+1)(2x-7)
f)x4+36=x4+12x2+36-12x2=(x2+6)2-12x2=(x2-\(\sqrt{12}x\)+6)(x2+\(\sqrt{12}x\)+6)
h)x4+4y4=x4+4x2y2+4y2-4x2y2=(x2+2y2)-4x2y2=(x2+2y2-2xy)(x2+2y2+2xy)
Từ \(\frac{x}{2}=\frac{y}{3}\) \(\Rightarrow\) \(\frac{3x}{2.3}=\frac{4y}{3.4}=\frac{3x}{6}=\frac{4y}{12}\) và 3x+4y=36
Theo tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{3x}{6}=\frac{4y}{12}=\frac{3x+4y}{6+12}=\frac{36}{18}=2\)
\(\frac{x}{2}=2\Rightarrow x=2.2=4;\frac{y}{3}=2\Rightarrow y=3.2=6\)
Vậy x=4 ; y=6
3x=2y
4x=5z
Ta có : 3x=2y <=> \(\frac{x}{2}=\frac{y}{3}\) <=> \(\frac{x}{10}=\frac{y}{15}\)
<=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{8}\)
4x=5z <=> \(\frac{x}{5}=\frac{z}{4}\)<=> \(\frac{x}{15}=\frac{z}{8}\)
=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{8}=\frac{x-y+z}{10-15+8}=\frac{36}{3}=12\)
Từ đó ta có :
\(\frac{x}{10}=12\Leftrightarrow x=120\)
\(\frac{y}{15}=12\Leftrightarrow y=180\)
\(\frac{z}{8}=12\Leftrightarrow z=96\)
Vậy x = 120 , y=180 , z=96
Chúc bạn học tốt ><
a, \(x^2-4x-5\)
\(=x^2-5x+x-5=x\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
b, \(x^2-4xy+4y^2-36\)
\(=\left(x-2y\right)^2-36=\left(x-2y-6\right)\left(x-2y+6\right)\)
c, \(x^2-9x+20=x^2-5x-4x+20\)
\(=x\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x-4\right)\left(x-5\right)\)
d, \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)\)
\(=\left(3x-6\right)\left(3x+6\right)-\left(3x-6\right)\left(2x+5\right)\)
\(=\left(3x-6\right)\left(3x+6-2x-5\right)\)
\(=3\left(x-2\right)\left(x+1\right)\)
a) \(x^2-4x-5=x^2-4x+4-9\)
= \(\left(x-2\right)^2-3^2\)
= \(\left(x-2-3\right)\left(x-2+3\right)=\left(x-5\right)\left(x+1\right)\)
b) \(x^2-4xy+4y^2-36\)
= \(\left(x-2y\right)^2-6^2=\left(x-2y-6\right)\left(x-2y+6\right)\)
c) \(x^2-9x+20\)
= \(\left(x-5\right)\left(x-4\right)\)
d) \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)\)
= \(\left(3x-6\right)\left(3x+6\right)-\left(3x-6\right)\left(2x+5\right)\)
= \(\left(3x-6\right)\left(3x+6-2x-5\right)=\left(3x-6\right)\left(x+1\right)\)
Đường thẳng song song d nên nhận (3;-4) là 1 vtpt
Phương trình:
\(3\left(x-2\right)-4\left(y-1\right)=0\Leftrightarrow3x-4y-2=0\)
1.\(x^2-2x-4y^2-4y=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
2.\(x^4+2x^3-4x-4=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)=\left(x^2-2\right)\left(x^2+2x-2\right)\)
3.\(3x^2-3y^2-2\left(x-y\right)^2=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\left(x-y\right)=\left(x-y\right)\left(3x+3y-2x+2y\right)\)\(=\left(x-y\right)\left(x+5y\right)\)
4.\(x^3-4x^2-9x+36=x^2\left(x-4\right)-9\left(x-4\right)=\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
5.\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
6.\(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)\(=\left(2x+1\right)\left(3-2x-5\right)=\left(2x+1\right)\left(-2-2x\right)=-2\left(2x+1\right)\left(x+1\right)\)
7.\(\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)\(=\left(x-5\right)\left(4x+1\right)\)
8.\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)=\left(3x-2\right)\left(3x-6\right)=3\left(3x-2\right)\left(x-2\right)\)
3x + 4y = 36
3x = 36 : 4 : y = 9 : y
x = 9 : y : 3 = 9 : 3 : y
x = 3 : y
=> y ϵ Ư(3) = {1;3;-1;-3}
Ta có bảng giá trị sau:
Vậy:
Đề bài là tìm x hay tìm y thế em