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c: \(=3.7\cdot12+3.7\cdot70+3.7\cdot18\)
=3,7(12+70+18)
=3,7*100=370
d: \(=202,2\left(3+6+1\right)=202,2\cdot10=2022\)
e: \(=4,5\cdot15+4,5\cdot90+4,5\cdot25=4,5\left(15+90+25\right)=585\)
i: \(=3.18\cdot4+3.18\cdot75+3.1\cdot8+3.18\cdot13\)
\(=3,18\left(4+75+13\right)+24,8\)
\(=317.36\)
c: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
d) Trạng ngữ : Buổi sáng hôm ấy , đã quá giờ đến lớp
e) Trên triền đê , chiều chiều
g) Đứng bên đó
d) Buổi sáng hôm ấy, đã quá giờ đến lớp
e) Đám trẻ mục đồng chúng tôi
g) Đứng bên đó
f: 5y^2-20=0
=>y^2-4=0
=>(y-2)(y+2)=0
=>y=2 hoặc y=-2
g: |x-2|-1=0
=>|x-2|=1
=>x-2=1 hoặc x-2=-1
=>x=3 hoặc x=1
g) \(\left\{{}\begin{matrix}x-1\ge0\\2-\sqrt{x-1}\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\sqrt{x-1}\le2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-4\le x-1\le4\end{matrix}\right.\)
\(\Leftrightarrow1\le x\le5\)
h) \(\left\{{}\begin{matrix}\dfrac{2x-4}{5-x}\ge0\\5-x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-4\ge0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-4\le0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\\x\ne5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}5>x\ge2\left(tm\right)\\5< x\le2\left(vl\right)\end{matrix}\right.\\x\ne5\end{matrix}\right.\)
\(\Leftrightarrow5>x\ge2\)
i) \(x^2-8x-9\ge0\)\(\Leftrightarrow\left(x-4\right)^2-25\ge0\Leftrightarrow\left(x-4\right)^2\ge25\)
\(\Leftrightarrow-5\ge x-4\ge5\)\(\Leftrightarrow-1\ge x\ge9\)
j) \(2x-x^2>0\)
\(\Leftrightarrow\left(x-1\right)^2< 1\)
\(\Leftrightarrow-1< x-1< 1\Leftrightarrow0< x< 2\)
a: ĐKXĐ: \(2\le x\le4\)
b: ĐKXĐ: x>0
c: ĐKXĐ: \(x< \dfrac{1}{3}\)
\(j,\left(\dfrac{-1}{2}\right)^3:1\dfrac{3}{8}-25\%\left(-6\dfrac{2}{11}\right)\)
\(=\dfrac{-1}{8}:\dfrac{11}{8}-\dfrac{1}{4}.\dfrac{-68}{11}\)
\(=\dfrac{-1}{11}-\dfrac{-17}{11}\)
\(=\dfrac{16}{11}\)
e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
e)x3-4x+14x(x-2)=0
⇔ x(x2-4)+14x(x-2)=0
⇔ x(x-2)(x+2)+14x(x-2)=0
⇔ (x-2)(x2+2x+14x)=0
⇔ x(x-2)(x+16)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)
g)x2(x+1)-x(x+1)+x(x-1)=0
⇔ (x+1)(x2-x)+x(x-1)=0
⇔ x(x+1)(x-1)+x(x-1)=0
⇔ x(x-1)(x+2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
\(g,=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+12-3\sqrt{3}\\ =\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)
pt<=>-cos4x-sin4x=sinx-cosx
<=>-(sin4x+cos4x)=sinx-cosx
<=>-\(\sqrt{2}\)sin(4x+\(\dfrac{\Pi}{4}\))=\(\sqrt{2}\)sin(x-\(\dfrac{\Pi}{4}\))
<=>sin(-4x-\(\dfrac{\Pi}{4}\))=sin(x-\(\dfrac{\Pi}{4}\))
tới đây bạn tự giải nha!!