\(tanx=tan20^0\)

\(\Rightarrow x=20^0+k180^0\) (\(k\in Z\))

=>sin 5x=sin(pi/2-x)

=>5x=pi/2-x+k2pi hoặc 5x=pi/2+x+k2pi

=>6x=pi/2+k2pi hoặc 4x=pi/2+k2pi

=>x=pi/12+kpi/3 hoặc x=pi/8+kpi/2

-2x+3=2

=>x=0,5

hoặc -2x+3=-2

=>x=2,5

Nguyễn Thị Phương Lan hình như mk lm đúng r đó ^^

hình như có 2 kết quả 

\(4FeS_2+11O_2\overset{t^o}{--->}2Fe_2O_3+8SO_2\)

Ta có: \(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\)

\(\Leftrightarrow\dfrac{-5}{\left(x-2\right)\left(x-3\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}=0\)

\(\Leftrightarrow-5-x^2+9=0\)

\(\Leftrightarrow x=-2\)

\(\sqrt{4x+1}+\sqrt{3x-2}=5\)

ĐKXĐ :\(\hept{\begin{cases}4x+1\ge0\\3x-2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-\frac{1}{4}\\x\ge\frac{2}{3}\end{cases}\Leftrightarrow x\ge\frac{2}{3}}}\)

Pt \(\Rightarrow\sqrt{4x+1}=5-\sqrt{3x-2}\)

\(\Leftrightarrow4x+1=\left(5-\sqrt{3x-2}\right)^2\)

\(\Leftrightarrow4x+1=25-10\sqrt{3x-2}+3x-2\)

\(\Leftrightarrow10\sqrt{3x-2}=-4x-1+25+3x-2\)

\(\Leftrightarrow10\sqrt{3x-2}=-x+22\)

\(\Leftrightarrow\left(10\sqrt{3x-2}\right)^2=\left(-x+22\right)^2\)

\(\Leftrightarrow100\left(3x-2\right)=484-44x+x^2\)

\(\Leftrightarrow300x-200=484-44x+x^2\)

\(\Leftrightarrow684-344x+x^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=342\\x=2\end{cases}}\)Tm

P/s ko bt có sai ở chỗ nào ko , bn tham khảo nha

\(m\left(mx-1\right)=\left(m+2\right)x-1\)

\(\Leftrightarrow m^2x-m=mx+2x-1\)

\(\Leftrightarrow m^2x-m-mx-2x+1=0\)

\(\Leftrightarrow mx\left(m-1\right)-\left(m-1\right)-2x=0\)

\(\left(mx-1\right)\left(m-1\right)-2x=0\)

tớ chỉ nghỉ ra có đến đó thôi

Ta có: \(\left\{{}\begin{matrix}x^3=3x+8y\\y^3=8x+3y\end{matrix}\right.\)

\(\Rightarrow x^3-y^3=5y-5x\)\(\Leftrightarrow x^3-y^3+5x-5y=0\)\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+5\right)=0\)

\(\Leftrightarrow x=y\)(vì \(x^2+xy+y^2+5>0\))

Thay \(x=y\) vào phương trình \(x^3=3x+8y\) ta được

\(x^3=11x\)\(\Leftrightarrow x\left(x^2-11\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=\sqrt{11}\\x=y=-\sqrt{11}\end{matrix}\right.\)