Sửa đề:

\((2x^2+x-2015)^2+4(x^2-5x-2016)^2=4(2x^2+x-2015)(x^2-5x-2016)\)

\(\Rightarrow\left(2x^2+x-2015\right)^2-2.\left(2x^2+x-2015\right).2.\left(x^2-5x-2016\right)+[2.\left(x^2-5x-2016\right)]^2=0\)

\(\Rightarrow[2x^2+x-2015-2.\left(x^2-5x-2016\right)]^2=0\)

\(\Rightarrow11x+2017=0\)

\(\Rightarrow x=\frac{-2017}{11}\)

\(\dfrac{x+2}{2016}+\dfrac{x+3}{2015}+\dfrac{x+4}{2014}+\dfrac{x+2036}{6}=0\)

<=>\(\dfrac{x+2}{2016}+1+\dfrac{x+3}{2015}+1+\dfrac{x+4}{2014}+1+\dfrac{x+2036}{6}-3=0\)

<=>\(\dfrac{x+2018}{2016}+\dfrac{x+2018}{2015}+\dfrac{x+2018}{2014}+\dfrac{x+2018}{6}=0\)

<=>\(\left(x+2018\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{6}\right)=0\)

vì 1/2016+1/2015+1/2014+1/6 khác 0

=>x+2018=0<=>x=-2018

vậy...................

chúc bạn học tốt ^ ^

a: ĐKXĐ: x>=0

b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)

\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)

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

\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)

\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)

=>\(x\in\left\{0;1.2996\right\}\)

pt <=> \(2x^2-20x+54-2\sqrt{x-4}-2\sqrt{6-x}=0\)

<=> \(\left(2x^2-20x+50\right)+\left(x-4-2\sqrt{x-4}+1\right)+\left(6-x-2\sqrt{6-x}+1\right)=0\)

<=> \(2\left(x-5\right)^2+\left(\sqrt{x-4}-1\right)^2+\left(\sqrt{6-x}-1\right)^2=0\)

<=> x = 5

1; Khi m=1 thì pt sẽ là \(\sqrt{x+1}=x+1\)

=>(x+1)^2=(x+1)

=>x(x+1)=0

=>x=0hoặc x=-1

2: \(\Leftrightarrow x+1=\left(x+m\right)^2\)

=>x^2+2mx+m^2-x-1=0

=>x^2+x(2m-1)+m^2-1=0

Δ=(2m-1)^2-4(m^2-1)

=4m^2-4m+1-4m^2+4

=-4m+5

Để pt có 2 nghiệm pb thì -4m+5>0

=>-4m>-5

=>m<5/4

Để pt có nghiệm kép thì 5-4m=0

=>m=5/4

Để pt vô nghiệm thì -4m+5<0

=>m>5/4

ĐKXĐ :  \(-4\le x\le4\)

TA CÓ : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)

\(\Leftrightarrow\left[\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\right]\left(\sqrt{4-x}+2\right)=2x\left(\sqrt{x+4}+2\right)\)

\(\Leftrightarrow\left[x+4-4\right]\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)

\(\Leftrightarrow x\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)

\(\Leftrightarrow x\left[\sqrt{4-x}+2-2\sqrt{x+4}-4\right]=0\)

\(\Leftrightarrow x=0\)HOẶC  \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)

VỚI \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)

\(\Leftrightarrow\sqrt{4-x}-2=2\sqrt{x+4}\)

\(\Leftrightarrow4-x+4-4\sqrt{4-x}=4x+16\)

\(\Leftrightarrow8-x-4x-16=4\sqrt{4-x}\)

\(\Leftrightarrow-5x-8=4\sqrt{4-x}\)ĐK : \(-4\le x\le\frac{-8}{5}\)

\(\Leftrightarrow\left[-\left(5x+8\right)\right]^2=16\left(4-x\right)\)

\(\Leftrightarrow25x^2+64+80x=64-16x\)

\(\Leftrightarrow25x^2+96x=0\Leftrightarrow x\left(25x+96\right)=0\)

\(\Leftrightarrow x=0\)HOẶC \(x=\frac{-96}{25}\)(THỎA MÃN ĐK )                                                                               

                                                                                               VẬY PT CÓ 2 NGHIỆM \(x\in\left[0;\frac{-96}{25}\right]\)

P/S : CÁCH CỦA MÌNH KHÁ DÀI VÀ CHI TIẾT QUÁ . BẠN CÓ THỂ THAM KHẢO CÁCH KHÁC NHANH HƠN :>

Lời giải:

Với mọi $x$ thuộc ĐKXĐ, ta luôn có:

\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}\geq 0\\ \sqrt{x^2+3x+1}\geq 0\end{matrix}\right.\)

Do đó, để \(\sqrt{3x+x^2+\frac{9}{4}}+\sqrt{x^2+3x+1}=0\) thì:

\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}= 0\\ \sqrt{x^2+3x+1}=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=\frac{-3}{2}\\ x=\frac{3\pm \sqrt{5}}{2}\end{matrix}\right.\) (vô lý)

Do đó pt vô nghiệm.