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Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) \(\frac{3x}{2x+4}+\frac{x+3}{x^2-4}\)
\(=\frac{3x}{2\left(x+2\right)}+\frac{x+3}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x\left(x-2\right)+2\left(x+3\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(=\frac{3x^2-6x+2x+6}{2\left(x^2-4\right)}\)
\(=\frac{3x^2-4x+6}{2\left(x^2-4\right)}\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)
\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)
\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)
chỗ cuối tớ sai
\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)
đây nha , e xin lỗi
a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))
\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)
\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)
\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)
\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)
\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)
b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)
\(=1+\dfrac{3}{x}\)
\(=\dfrac{x}{x}+\dfrac{3}{x}\)
\(=\dfrac{x+3}{x}\)
a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)
\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)
b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)
Bài 1:
a)
\(9x^2-49=0\)
\(9x^2-49+49=0+49.\)
\(9x^2=49\)
\(\frac{9x^2}{9}=\frac{49}{9}\)
\(x^2=\frac{49}{9}\)
\(x=\sqrt{\frac{49}{9}}\)
\(x=\frac{\sqrt{49}}{\sqrt{9}}\)
\(x=\frac{7}{3}\)hay \(x=2,33333...\)
b)
\(\left(x-1\right)\left(x+2\right)-x-2=0.\)
\(x^2+x-2-x-2.\)
\(x^2+\left(x-x\right)-\left(2+2\right)=\)\(0\)
\(x^2-4=0\)
\(x=\sqrt{4}\)
\(x=2\)
Bài 2:
a)
\(\frac{x}{x}-3+9-\frac{6x}{x^2}-3x.\)
\(=1-3+9-\frac{6x}{x^2}-3x.\)
\(=1-3+9-\frac{6}{x}-3x.\)
\(=7-\frac{6}{x}-3x\)
b)
\(6x-\frac{3}{x}\div4x^2-\frac{1}{3x^2}\)
\(=6x-\frac{3}{x}\div\frac{4}{1}x^2-\frac{1}{3x^2}.\)
\(=6x-\frac{3}{x}\times\frac{1}{4}x^2-\frac{1}{3x^2}\)
\(=6x-\frac{3x^2}{x4}-\frac{1}{3x^2}\)
\(=6x-\frac{3x}{4}-\frac{1}{3x^2}\)
\(=\frac{6x}{1}-\frac{3x}{4}-\frac{1}{3x^2}\)
\(=\frac{72x^3-36x^3-12x^2}{12x^2}\)
\(=\frac{36-12x^2}{12x^2}\)
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
mk ko biết làm
xin lỗi bn nhae
xin lỗi vì đã ko giúp được bn
chcus bn học gioi!
nhae@@@
a, 2x(1-3)=0
=>...................
hk tốt
a) \(2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)