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Câu a hạ bậc rồi áp dụng cosa + cosb
Câu b thì mối liên hệ giữa tan với cot là ra
Bài 1:
\(\left(x+4\right)\left(y+3\right)=3\)
\(\Rightarrow\left[{}\begin{matrix}x+4=3\\y+3=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-4\\y=3-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)
Vậy \(x=-1;y=0\)
b) \(\dfrac{4}{3}-\left(x-\dfrac{1}{5}\right)=\left|-\dfrac{3}{10}+\dfrac{1}{2}\right|-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{3}-x+\dfrac{1}{5}=\left|\dfrac{1}{5}\right|-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{4}{3}-x+\dfrac{1}{5}=\dfrac{1}{5}-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{4}{3}-x=-\dfrac{1}{6}\)
\(\Leftrightarrow-x=-\dfrac{1}{6}-\dfrac{4}{3}\)
\(\Leftrightarrow-x=-\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy \(x=\dfrac{3}{2}\)
\(=-6\cdot\dfrac{1}{27}\cdot\left[\dfrac{-4}{9}\cdot\left(\dfrac{-1}{2}-\dfrac{4}{3}\right)\right]\)
\(=\dfrac{-2}{9}\cdot\left[-\dfrac{4}{9}\cdot\dfrac{-11}{6}\right]\)
\(=\dfrac{-2}{9}\cdot\dfrac{44}{54}=\dfrac{-88}{432}=\dfrac{-11}{54}\)
Ta có: 2. |3x - 1| + 1 = 5
=> 2. |3x - 1| = 5 - 1 = 4
=> |3x - 1| = 4/2 = 2
=> 3x - 1 = 2 hoặc 3x - 1 = -2
+/ 3x - 1 = 2
=> 3x = 2 +1 = 3
=> x = 3/3 =1
+/ 3x - 1 = -2
=> 3x = -2 + 1 = -1
=> x = -1/3
Vậy x thuộc {1; -1/3}.
\(2\left|3x-1\right|+1=5\)
\(\Leftrightarrow2.\left|2x-1\right|=5-1\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{5-1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=1\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
1: ĐKXĐ: \(x^3-6x^2+11x-6< >0\)
\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6\ne0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)\ne0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\ne0\)
hay \(x\notin\left\{1;2;3\right\}\)
2; ĐKXĐ: \(\left\{{}\begin{matrix}3-2x>=0\\x+1< >-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\x< >-2\end{matrix}\right.\)
3: ĐKXĐ: \(\left\{{}\begin{matrix}x+2< >0\\x-1< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< >-2\\x< >1\end{matrix}\right.\Leftrightarrow x\in R\)
\(VT=\dfrac{1+cos2x}{cos2x}\times\dfrac{1+cos4x}{sin4x}\) (*)
Ta có: theo công thức hạ bậc có: \(cos^2x=\dfrac{1+cos2x}{2}\Leftrightarrow1+cos2x=2cos^2x\) (1)
Ta có: \(cos2x=1-sin^2x\Rightarrow cos4x=1-2sin^22x\) (2)
Tương Tự có \(sin2x=2sinx\times cosx\Rightarrow sin4x=2sin2x\times cos2x\) (3)
Thay (1),(2),(3) vào (*) ta được: \(VT=\dfrac{2cos^2x}{cos2x}\times\dfrac{1+\left(1-2sin^22x\right)}{2sin2x\times cos2x}\)
\(VT=\dfrac{2cos^2x\times2\left(1-sin^22x\right)}{cos^22x\times2sin2x}\) mà \(1-sin^22x=cos^22x\)
\(\Rightarrow VT=\dfrac{2cos^2x\times cos^22x}{cos^22x\times2sinx\times cosx}=\dfrac{cosx}{sinx}=tanx\left(đpcm\right)\)
đoạn cuối nhầm nha \(VT=\dfrac{cosx}{sinx}=cotx\left(đpcm\right)\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
a) ta có :
\(\Delta'=1^2-\left(-1-m\right)\left(m^2-1\right)=1-\left(-m^2+1-m^3+m\right)=1+m^2-1+m^3-m=m^3+m^2-m=m\left(m^2+m-1\right)\)để phương trình có nghiệm thì \(\Delta\ge0\)
hay \(m\left(m^2+m-1\right)\ge0\)
=> \(\left\{{}\begin{matrix}m\ge0\\m^2+m-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\\left(m+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}m\ge0\\\left(m+\dfrac{1}{2}\right)^2\ge\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}m\ge0\\\left[{}\begin{matrix}m+\dfrac{1}{2}\ge\\m+\dfrac{1}{2}\le-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\dfrac{\sqrt{5}}{2}}\)
What? Lớp 10? Mí bài nỳ dễ mak! Trên lp cs hc mak k giải đc thì thui lun!
tui mới lớp 7 mà