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\(\Leftrightarrow\dfrac{1-cos2x}{2}-\left(1+\sqrt{3}\right)\cdot\dfrac{1}{2}sin2x+\sqrt{3}\cdot\dfrac{1+cos2x}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x-\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot sin2x+\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}cos2x=0\)
\(\Leftrightarrow cos2x\left(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot sin2x=\dfrac{-\sqrt{3}-1}{2}\)
\(\Leftrightarrow sin2x\cdot\dfrac{-\sqrt{3}-1}{2}+cos2x\cdot\dfrac{\sqrt{3}-1}{2}=\dfrac{-\sqrt{3}-1}{2}\)
\(\Leftrightarrow sin2x\left(-\sqrt{3}-1\right)+cos2x\left(\sqrt{3}-1\right)=-\sqrt{3}-1\)
\(\Leftrightarrow sin2x\cdot\dfrac{-\sqrt{3}-1}{8}+cos2x\cdot\dfrac{\sqrt{3}-1}{8}=\dfrac{-\sqrt{3}-1}{8}\)
\(\Leftrightarrow sin\left(2x+a\right)=cosa=sin\left(\dfrac{pi}{2}-a\right)\)(với \(cosa=\dfrac{-\sqrt{3}-1}{8}\))
\(\Leftrightarrow\left[{}\begin{matrix}2x+a=\dfrac{pi}{2}-a+k2pi\\2x+a=pi-\dfrac{pi}{2}+a+k2pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-2a+\dfrac{pi}{2}+k2pi\\2x=\dfrac{pi}{2}+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-a+\dfrac{pi}{4}+kpi\\x=\dfrac{pi}{4}+kpi\end{matrix}\right.\)
a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow x=30^o\)
b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)
\(\Leftrightarrow x=30^o\)
c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)
d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)
Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(
e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)
f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)
1: \(=\dfrac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}\)
\(=\dfrac{\dfrac{1}{cotx}+cotx+2}{2+tanx+cotx}\)
\(=1\)
2: \(VT=\dfrac{cos^2x+cosxsinx+sin^2x-sinx\cdot cosx}{sin^2x-cos^2x}\)
\(=\dfrac{1}{sin^2x-cos^2x}\)
\(VP=\dfrac{1+cot^2x}{1-cot^2x}=\left(1+\dfrac{cos^2x}{sin^2x}\right):\left(1-\dfrac{cos^2x}{sin^2x}\right)\)
\(=\dfrac{1}{sin^2x}:\dfrac{sin^2x-cos^2x}{sin^2x}=\dfrac{1}{sin^2x-cos^2x}\)
=>VT=VP
\(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\)
\(\Leftrightarrow11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{11x-5-2x^2}\)
\(\Leftrightarrow121\left(5-x\right)+176\sqrt{\left(5-x\right)\left(2x-1\right)}+64\left(2x-1\right)=576+144\sqrt{11x-5-2x^2}\)\(+9\left(11x-5-2x^2\right)\)
\(\Leftrightarrow605-121x+176\sqrt{11x-5-2x^2}+128x-64=576+144\sqrt{11x-5-2x^2}\)\(+99x-18x^2\)
\(\Leftrightarrow176\sqrt{11x-5-2x^2}-144\sqrt{11x-5-2x^2}=531+99x-18x^2-541-7x\)
\(\Leftrightarrow32\sqrt{11x-5-2x^2}=-10+92x-18x^2\)
\(\Leftrightarrow16\sqrt{11x-5-2x^2}=-5+46x-9x^2\)
\(\Leftrightarrow256\left(11x-5-2x^2\right)=25+2116x^2+81x^4-460x+90x^2-823x^3\)
\(\Leftrightarrow2816x-1280-512x^2=25+2206x^2+81x^4-460x-823x^3\)
\(\Leftrightarrow9\left(364x-145-302x^2-9x^4+92x^3\right)=0\)
\(\Leftrightarrow-9x^4+92x^3-302x^2+364x-145=0\)
\(\Leftrightarrow-\left(x-1\right)\left(9x^3-83x^2+219x-145\right)=0\)
\(\Leftrightarrow-\left(x-1\right)\left(x-1\right)\left(9x^2-74x+145\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\left(9x-29\right)\left(x-5\right)=0\Leftrightarrow\)x=1; x=29/9; x=5
\(\Leftrightarrow11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{11x-5-2x^2}\)
a/
ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
\(\Rightarrow x=3\)
b/ \(\left\{{}\begin{matrix}2x-y\ge1\\x+2y\ge0\end{matrix}\right.\) (1)
Biến đổi pt dưới:
\(\left(2\left(x+2y\right)-1\right)\sqrt{2x-y-1}=\left(2\left(2x-y-1\right)-1\right)\sqrt{x+2y}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2y}=a\ge0\\\sqrt{2x-y-1}=b\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(2a^2-1\right)b=\left(2b^2-1\right)a\)
\(\Leftrightarrow2a^2b-2ab^2+a-b=0\)
\(\Leftrightarrow2ab\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2ab+1\right)=0\)
\(\Rightarrow a=b\) (do \(\left\{{}\begin{matrix}a\ge0\\b\ge0\end{matrix}\right.\) \(\Rightarrow2ab+1>0\))
\(\Rightarrow\sqrt{x+2y}=\sqrt{2x-y-1}\Leftrightarrow x+2y=2x-y-1\)
\(\Leftrightarrow x=3y+1\)
Thế vào pt trên:
\(\left(3y+1\right)^2-5y^2-8y-3=0\)
\(\Leftrightarrow4y^2-2y-2=0\) \(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=4\\y=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}\end{matrix}\right.\)
Thế nghiệm vào hệ điều kiện (1) thì chỉ có \(\left(x;y\right)=\left(4;1\right)\) thỏa mãn
Câu a) Cứ bình phương và bình phương cho hết căn rồi bấm máy tính giải ra :v
b)pt\(\left(2\right)\)\(\Leftrightarrow\left(2x+4y-1\right)^2\left(2x-y-1\right)=\left(4x-2y-3\right)^2\left(x+2y\right)\)
\(\Leftrightarrow\left(x-3y-1\right)\left(8x^2-8y^2-4x-8y+12xy-1\right)=0\)
Đến đây tự giải thế vào (1)
Nguyễn Việt Lâm Giải giúp t TH2 nha!
cái này trên OLM mà
cái này chắc cũng lớp 10 chứ ko thoát đâu