Giải phương trình sau:
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2)\(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
⇔\(\dfrac{x+5}{3\left(x-2\right)}-\dfrac{1}{2}=\dfrac{2x-3}{2\left(x-2\right)}\)
⇔\(\dfrac{2\left(x+5\right)}{3.2\left(x-2\right)}-\dfrac{3\left(x-2\right)}{2.3\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{2.3\left(x-2\right)}\)
⇔\(\dfrac{2x+10}{6\left(x-2\right)}-\dfrac{3x-6}{6\left(x-2\right)}=\dfrac{6x-9}{6\left(x-2\right)}\)
⇔\(2x+10-\left(3x-6\right)=6x-9\)
⇔\(2x+10-3x+6-6x+9=0\)
⇔\(-7x+25=0\)
⇔\(-7x=-25\)
⇔\(x=\dfrac{25}{7}\)
3)\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)
⇔\(\dfrac{1}{x-1}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)
⇔\(\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇔\(x^2+x+1-3x^2=2x^2-2x\)
⇔\(x^2+x+1-3x^2-2x^2+2x=0\)
⇔\(-4x^2+3x+1=0\)
⇔\(-\left(4x^2-3x-1\right)=0\)
⇔\(-\left(4x^2-4x+x-1\right)=0\)
⇔\(-\left[\left(4x^2-4x\right)+\left(x-1\right)\right]=0\)
⇔\(-\left[4x\left(x-1\right)+\left(x-1\right)\right]=0\)
⇔\(-\left[\left(x-1\right)\left(4x+1\right)\right]=0\)
⇔\(-\left(x-1\right)=0\) hay \(-\left(4x+1\right)=0\)
⇔\(-x+1=0\) hay \(-4x-1=0\)
⇔\(-x=-1\) hay \(-4x=1\)
⇔\(x=1\) hay \(x=-\dfrac{1}{4}\)
thể tích là
50x15x40=3000(dm3)
3000dm3=3m3=3 khối
Đ/S:3 khối cát
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{2}\right)=\dfrac{1}{2}+\dfrac{1}{2}cos2x\)
\(\Leftrightarrow-sin2x=cos2x\)
\(\Leftrightarrow sin2x+cos2x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow2x+\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\)
Cách biến đổi (chứng minh) đơn giản thôi:
\(sin2x+cos2x=0\Leftrightarrow\sqrt{2}\left(\dfrac{\sqrt{2}}{2}sin2x+\dfrac{\sqrt{2}}{2}cos2x\right)=0\)
\(\Leftrightarrow\sqrt{2}\left[sin2x.cos\left(\dfrac{\pi}{4}\right)+cos2x.sin\left(\dfrac{\pi}{4}\right)\right]=0\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=0\)
Hoặc bạn nhớ vài công thức hay gặp sau:
\(sina+cosa=\sqrt{2}sin\left(a+\dfrac{\pi}{4}\right)=\sqrt{2}cos\left(a-\dfrac{\pi}{4}\right)\)
\(sina-cosa=\sqrt{2}sin\left(a-\dfrac{\pi}{4}\right)=-\sqrt{2}cos\left(a+\dfrac{\pi}{4}\right)\)