a, Với \(x>0;x\ne4;x\ne9\)

\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)

b, Ta có : A = -2 hay 

\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)

\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)

\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)

bình phương 2 vế ta có : 

\(x=\left(2x+3\right)^2=4x^2+12x+9\)

\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)

\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)              

\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)

\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)

\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)

\(=\frac{4x}{\sqrt{x}-3}\)

\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right)\div\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right)\div\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\div\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{4\sqrt{x}}{2-\sqrt{x}}\times\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}=\frac{4\sqrt{x}}{2-\sqrt{x}}\times\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\frac{4\sqrt{x}}{\sqrt{x}-3}\)

Để A = -2 thì \(\frac{4\sqrt{x}}{\sqrt{x}-3}=-2\)

=> \(4\sqrt{x}=-2\sqrt{x}+6\)

<=> \(6\sqrt{x}=6\)<=> \(\sqrt{x}=1\)<=> x = 1 ( tm )

Vậy với x = 1 thì A = -2

Bạn ơi bạn đã giải được bài 1 chưa vậy? 

Câu2 : <=> 2x⁴-2x²+5x⁵-5=0
<=>2x²(x²-1)+5(x²-1)=0

<=>(2x²+5)(x-1)(x+1)=0

<=> x={+-1 } vì 2x²+5>0 mọi x

Câu 2:<=>3x³-3x²+13x²-13x=0 <=> 3x²(x-1)+13x(x-1)=0 <=> x(3x²+13)(x-1)=0 <=>x={0;1) vì 3x²+13>0 mọi X

Xin lỗi mng là câu 1D là

Luôn đúng với mọi m thuộc R nha mng.

Dễ thấy \(x=2017\)không là nghiệm của phương trình.

Ta có:

\(\frac{1+\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)^2}{1-\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)}=\frac{13}{37}\)

Đặt \(\frac{x-2018}{2017-x}=a\)

\(\Rightarrow\frac{1+a+a^2}{1-a+a^2}=\frac{13}{37}\)

\(\Leftrightarrow24a^2+50a+24=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\\a=-\frac{4}{3}\end{cases}}\)

a.

\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)

c.

\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)

d.

\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)

Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì

\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)

Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)

Vậy \(\left(a;b\right)=\left(9;4\right)\)

\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)

Có bài ngược của bài này, bạn đăng và đã có lời giải thì chỉ cần đảo lại đáp án là được.

\(E=\sqrt{x}+\dfrac{4}{\sqrt{x}}-2=\dfrac{4\sqrt{x}}{9}+\dfrac{4}{\sqrt{x}}+\dfrac{5}{9}.\sqrt{x}-2\)

\(E\ge2\sqrt{\dfrac{16\sqrt{x}}{9\sqrt{x}}}+\dfrac{5}{9}.\sqrt{9}-2=\dfrac{7}{3}\)

\(E_{min}=\dfrac{7}{3}\) khi \(x=9\)

\(F=3\sqrt{x}+\dfrac{1}{\sqrt{x}}+1=2\sqrt{x}+\dfrac{1}{\sqrt{x}}+\sqrt{x}+1\)

\(F\ge2\sqrt{\dfrac{2\sqrt{x}}{\sqrt{x}}}+1.\sqrt{\dfrac{1}{2}}+1=\dfrac{2+5\sqrt{2}}{2}\)

\(F_{min}=\dfrac{2+5\sqrt{2}}{2}\) khi \(x=\dfrac{1}{2}\)