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Bài 1:
a: \(\frac{y-1}{y+1}=\frac{\left.\left(y-1\right)\left(y-1\right)\right.}{\left(y+1\right)\left(y-1\right)}=\frac{\left(y-1\right)^2}{\left(y+1\right)\left(y-1\right)}\)
\(\frac{y+1}{y-1}=\frac{\left(y+1\right)\left(y+1\right)}{\left(y-1\right)\left(y+1\right)}=\frac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)}\)
\(\frac{1}{y^2-1}=\frac{1}{\left(y-1\right)\left(y+1\right)}\)
b: \(\frac{2}{y^2-4y}=\frac{2}{y\left(y-4\right)}=\frac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)}=\frac{2y+8}{y\left(y-4\right)\left(y+4\right)}\)
\(\frac{y}{y^2-16}=\frac{y}{\left(y-4\right)\left(y+4\right)}=\frac{y\cdot y}{y\left(y-4\right)\left(y+4\right)}=\frac{y^2}{y\left(y-4\right)\left(y+4\right)}\)
c: \(\frac{x}{x^3+1}=\frac{x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{x-1}{x+1}=\frac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{x+2}{x^2-x+1}=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\)
d: \(\frac{2}{x^2+5x}=\frac{2}{x\left(x+5\right)}=\frac{2\left(x+5\right)}{x\left(x+5\right)^2}=\frac{2x+10}{x\left(x+5\right)^2}\)
\(\frac{x+5}{x^2+10x+25}=\frac{x+5}{\left.\left(x+5\right)^2\right.}=\frac{x\left(x+5\right)}{x\left(x+5\right)^2}\)
\(\frac{x+2}{x}=\frac{\left(x+2\right)\left(x+5\right)^2}{x\left(x+5\right)^2}\)
e: \(\frac{2x-3xy}{2xy}=\frac{x\left(2-3y\right)}{2xy}\)
\(\frac{x+2y}{x}=\frac{2y\left(x+2y\right)}{2xy}=\frac{2xy+4y^2}{2xy}\)
\(\frac{x-5y}{y}=\frac{2x\left(x-5y\right)}{2x\cdot y}=\frac{2x^2-10xy}{2xy}\)
f: \(\frac{x}{x^3-xy^2}=\frac{x}{x\left(x^2-y^2\right)}=\frac{1}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)
\(\frac{1}{\left(x+y\right)^2}=\frac{1\cdot\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}=\frac{\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}\)
\(\frac{1}{\left(x-y\right)^2}=\frac{1\cdot\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}=\frac{\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)
g: \(\frac{x}{x^3-4x}=\frac{x}{x\left(x^2-4\right)}=\frac{1}{x^2-4}=\frac{1}{\left(x-2\right)\left(x+2\right)}\)
\(\frac{5}{2-x}=\frac{-5}{x-2}=\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-5x-10}{\left(x-2\right)\left(x+2\right)}\)
\(\frac{3x+1}{x+2}=\frac{\left(3x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x^2-5x-2}{\left(x+2\right)\left(x-2\right)}\)
h: \(\frac{3x}{x^3-1}=\frac{3x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{x+2}{x-1}=\frac{\left(x+2\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{3-x}{x^2+x+1}=\frac{\left(3-x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
i: \(\frac{3}{x^2+6x+9}=\frac{3}{\left(x+3\right)^2}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)^2}=\frac{3x-9}{\left(x-3\right)\left(x+3\right)^2}\)
\(\frac{x-1}{x+3}=\frac{\left(x-1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x_{}^2-9\right)}{\left(x-3\right)\left(x+3\right)^2}\)
\(\frac{2x-6}{x^2-9}=\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)^2}\)
mọi người giải giúp em với ạ em đang cần gấp lắm ạ
Mọi Người giải giúp em ạ em cảm ơn ạ https://classroom.google.com/c/MzgyNzMxODQzMzkw/a/NDM1MTIwOTcyNDIz/details
giúp em bài 1 với ạ( link)
giúp mk nha mờn nhìu ạk
Bài 1:
Vận tốc cano khi dòng nước lặng là: $25-2=23$ (km/h)
Bài 2:
Đổi 1 giờ 48 phút = 1,8 giờ
Độ dài quãng đường AB: $1,8\times 25=45$ (km)
Vận tốc ngược dòng là: $25-2,5-2,5=20$ (km/h)
Cano ngược dòng từ B về A hết:
$45:20=2,25$ giờ = 2 giờ 15 phút.
Giúp em với:((((
\(2x+3y+5z=\frac{x^2+y^2+z^2}{2}+19\)
\(x^2+y^2+z^2+38=4x+6y+10z\)
\(\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\left(x-2\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
\(x-2=y-3=z-5=0\)
\(x=2,y=3,z=5\)
Bài 1:
a.
$a^3-a^2c+a^2b-abc=a^2(a-c)+ab(a-c)$
$=(a-c)(a^2+ab)=(a-c)a(a+b)=a(a-c)(a+b)$
b.
$(x^2+1)^2-4x^2=(x^2+1)^2-(2x)^2=(x^2+1-2x)(x^2+1+2x)$
$=(x-1)^2(x+1)^2$
c.
$x^2-10x-9y^2+25=(x^2-10x+25)-9y^2$
$=(x-5)^2-(3y)^2=(x-5-3y)(x-5+3y)$
d.
$4x^2-36x+56=4(x^2-9x+14)=4(x^2-2x-7x+14)$
$=4[x(x-2)-7(x-2)]=4(x-2)(x-7)$
Bài 2:
a. $(3x+4)^2-(3x-1)(3x+1)=49$
$\Leftrightarrow (3x+4)^2-[(3x)^2-1]=49$
$\Leftrightarrow (3x+4)^2-(3x)^2=48$
$\Leftrightarrow (3x+4-3x)(3x+4+3x)=48$
$\Leftrightarrow 4(6x+4)=48$
$\Leftrightarrow 6x+4=12$
$\Leftrightarrow 6x=8$
$\Leftrightarrow x=\frac{4}{3}$
b. $x^2-4x+4=9(x-2)$
$\Leftrightarrow (x-2)^2=9(x-2)$
$\Leftrightarrow (x-2)(x-2-9)=0$
$\Leftrightarrow (x-2)(x-11)=0$
$\Leftrightarrow x-2=0$ hoặc $x-11=0$
$\Leftrightarrow x=2$ hoặc $x=11$
c.
$x^2-25=3x-15$
$\Leftrightarrow (x-5)(x+5)=3(x-5)$
$\Leftrightarrow (x-5)(x+5-3)=0$
$\Leftrightarrow (x-5)(x+2)=0$
$\Leftrightarrow x-5=0$ hoặc $x+2=0$
$\Leftrightarrow x=5$ hoặc $x=-2$
Bài 1:
a: \(\frac{y-1}{y+1}=\frac{\left.\left(y-1\right)\left(y-1\right)\right.}{\left(y+1\right)\left(y-1\right)}=\frac{\left(y-1\right)^2}{\left(y+1\right)\left(y-1\right)}\)
\(\frac{y+1}{y-1}=\frac{\left(y+1\right)\left(y+1\right)}{\left(y-1\right)\left(y+1\right)}=\frac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)}\)
\(\frac{1}{y^2-1}=\frac{1}{\left(y-1\right)\left(y+1\right)}\)
b: \(\frac{2}{y^2-4y}=\frac{2}{y\left(y-4\right)}=\frac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)}=\frac{2y+8}{y\left(y-4\right)\left(y+4\right)}\)
\(\frac{y}{y^2-16}=\frac{y}{\left(y-4\right)\left(y+4\right)}=\frac{y\cdot y}{y\left(y-4\right)\left(y+4\right)}=\frac{y^2}{y\left(y-4\right)\left(y+4\right)}\)
c: \(\frac{x}{x^3+1}=\frac{x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{x-1}{x+1}=\frac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{x+2}{x^2-x+1}=\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+3x+2}{\left(x+1\right)\left(x^2-x+1\right)}\)
d: \(\frac{2}{x^2+5x}=\frac{2}{x\left(x+5\right)}=\frac{2\left(x+5\right)}{x\left(x+5\right)^2}=\frac{2x+10}{x\left(x+5\right)^2}\)
\(\frac{x+5}{x^2+10x+25}=\frac{x+5}{\left.\left(x+5\right)^2\right.}=\frac{x\left(x+5\right)}{x\left(x+5\right)^2}\)
\(\frac{x+2}{x}=\frac{\left(x+2\right)\left(x+5\right)^2}{x\left(x+5\right)^2}\)
e: \(\frac{2x-3xy}{2xy}=\frac{x\left(2-3y\right)}{2xy}\)
\(\frac{x+2y}{x}=\frac{2y\left(x+2y\right)}{2xy}=\frac{2xy+4y^2}{2xy}\)
\(\frac{x-5y}{y}=\frac{2x\left(x-5y\right)}{2x\cdot y}=\frac{2x^2-10xy}{2xy}\)
f: \(\frac{x}{x^3-xy^2}=\frac{x}{x\left(x^2-y^2\right)}=\frac{1}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)
\(\frac{1}{\left(x+y\right)^2}=\frac{1\cdot\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}=\frac{\left(x-y\right)^2}{\left(x+y\right)^2\cdot\left(x-y\right)^2}\)
\(\frac{1}{\left(x-y\right)^2}=\frac{1\cdot\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}=\frac{\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)
g: \(\frac{x}{x^3-4x}=\frac{x}{x\left(x^2-4\right)}=\frac{1}{x^2-4}=\frac{1}{\left(x-2\right)\left(x+2\right)}\)
\(\frac{5}{2-x}=\frac{-5}{x-2}=\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-5x-10}{\left(x-2\right)\left(x+2\right)}\)
\(\frac{3x+1}{x+2}=\frac{\left(3x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x^2-5x-2}{\left(x+2\right)\left(x-2\right)}\)
h: \(\frac{3x}{x^3-1}=\frac{3x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{x+2}{x-1}=\frac{\left(x+2\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{3-x}{x^2+x+1}=\frac{\left(3-x\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
i: \(\frac{3}{x^2+6x+9}=\frac{3}{\left(x+3\right)^2}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)^2}=\frac{3x-9}{\left(x-3\right)\left(x+3\right)^2}\)
\(\frac{x-1}{x+3}=\frac{\left(x-1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x_{}^2-9\right)}{\left(x-3\right)\left(x+3\right)^2}\)
\(\frac{2x-6}{x^2-9}=\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)^2}\)