À câu a mình tự làm được rồi nhé! Các bạn chỉ cần làm câu b cho mình là được.

b, \(\frac{2\sqrt{x}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\)

ĐK \(x\ge0\)

Pt 

<=> \(2\sqrt{x}+\sqrt{x\left(x+1\right)}=\sqrt{\left(x+1\right)\left(x+9\right)}\)

<=> \(4x+x^2+x+4\sqrt{x^2\left(x+1\right)}=x^2+10x+9\)

 <=> \(4x\sqrt{x+1}=5x+9\)

<=> \(16x^2\left(x+1\right)=25x^2+90x+81\)với mọi \(x\ge0\)

<=> \(16x^3-9x^2-90x-81=0\)

<=> \(x=3\)(tm ĐK)

Vậy x=3

a/ \(\hept{\begin{cases}VT=\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge2+3=5\\VP=4-2x-x^2=5-\left(x+1\right)^2\le5\end{cases}}\)

Dấu = xảy ra khi \(x=-1\)

b/ \(\sqrt{x-2}+\sqrt{4-x}=x^2-6x+11\)

Đặt \(\hept{\begin{cases}\sqrt{x-2}=a\ge0\\\sqrt{4-x}=b\ge0\end{cases}}\)thì ta có

\(\hept{\begin{cases}a^2+b^2=2\\a+b=-a^2b^2+3\end{cases}}\)

Đặt \(\hept{\begin{cases}a+b=S\\ab=P\end{cases}}\) thì ta có

\(\hept{\begin{cases}S^2-2P=2\\S=3-P^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(3-P^2\right)^2-2P=2\\S=3-P^2\end{cases}}\)

Thôi làm tiếp đi làm biếng quá.

a)√3x2+6x+7+√5x2+10x+14=4−2x−x2

\(\Leftrightarrow16x+\left(\sqrt{6}+\sqrt{10}\right)\sqrt{x}+21\)

\(\Leftrightarrow-x^2-2x+4\)

  Thế vào ta được:

\(x^2+18x+\left(\sqrt{6}+\sqrt{10}\right)\sqrt{x}=-17\)

\(x^2+18x+\left(\sqrt{6}+\sqrt{10}\right)\sqrt{x}+17=0\)

\(16x+\left(\sqrt{6}+\sqrt{10}\right)\sqrt{x}+21=4-x\left(x+2\right)\)

b)√x−2+√4−x=x2−6x+11

\(\Leftrightarrow\sqrt{x}-x=x^2-6x+11\)

\(\Leftrightarrow\sqrt{x}-x\)

\(\Leftrightarrow x^2-6x+11\)

\(\Leftrightarrow-x^2+5x+\sqrt{x}=11\)

\(\Leftrightarrow\sqrt{x}-x=\left(x-6\right)x+11\)

\(\Leftrightarrow-\left(\sqrt{x}-1\right)\sqrt{x}=x^2-6x+11\)

  Tới đây thì đơn giản rồi nhé!

a)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

\(pt\Leftrightarrow\sqrt{3x^2+6x+3+4}+\sqrt{5x^2+10x+5+9}=-x^2-2x+4\)

\(\Leftrightarrow\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}=-x^2-2x+4\)

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-x^2-2x+4\)

Dễ thấy: \(\hept{\begin{cases}3\left(x+1\right)^2\ge0\\5\left(x+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3\left(x+1\right)^2+4\ge4\\5\left(x+1\right)^2+9\ge9\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{3\left(x+1\right)^2+4}\ge2\\\sqrt{5\left(x+1\right)^2+9}\ge3\end{cases}}\)

\(\Rightarrow VT=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge2+3=5\)

Và \(VP=-x^2-2x+4=-x^2-2x-1+5\)

\(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\)

SUy ra \(VT\ge VP=5\Leftrightarrow x=-1\)

b)\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

\(pt\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1}=1\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2-\sqrt{x-1}=1\)

..... giải nốt tiếp ra x=1

c)Sửa đề \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)

ĐK:....

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\)

\(\le\left(1+1\right)\left(x-7+9-x\right)=4\)

\(\Rightarrow VT^2\le4\Rightarrow VT\le2\)

Lại có: \(VP=x^2-16x+66=x^2-16x+64+2\)

\(=\left(x-8\right)^2+2\ge2\)

Suy ra \(VT\ge VP=2\) khi \(VT=VP=2\)

\(\Rightarrow\left(x-8\right)^2+2=2\Rightarrow x-8=0\Rightarrow x=8\)

ĐKXĐ: \(x\in R\)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

=>\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x-4=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x+1-5=0\)

=>\(\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+14}-3+\left(x+1\right)^2=0\)

=>\(\dfrac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+14-9}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>

\(\dfrac{3x^2+6x+3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+5}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\dfrac{3\left(x^2+2x+1\right)}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x^2+2x+1\right)}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5}{\sqrt{5x^2+10x+14}+3}+1\right)=0\)

=>\(\left(x+1\right)^2=0\)

=>x+1=0

=>x=-1(nhận)

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

Câu a:

ĐKXĐ: \(x\geq 1\)

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

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

\(\Rightarrow x-1=8x-3+2\sqrt{(3x-2)(5x-1)}\) (bình phương 2 vế)

\(\Leftrightarrow 7x-2+2\sqrt{(3x-2)(5x-1)}=0\)

(Vô lý với mọi \(x\geq 1\) )

Do đó PT vô nghiệm.

Câu b)

PT \(\Leftrightarrow \sqrt{3(x^2+2x+1)+4}+\sqrt{5(x^2+2x+1)+9}=5-(x^2+2x+1)\)

\(\Leftrightarrow \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}=5-(x+1)^2\)

Vì \((x+1)^2\geq 0, \forall x\) nên:

\(\sqrt{3(x+1)^2+4}\geq \sqrt{4}=2\)

\(\sqrt{5(x+1)^2+9}\geq \sqrt{9}=3\)

\(\Rightarrow \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}\geq 5(1)\)

Mặt khác ta cũng có: \(5-(x+1)^2\leq 5-0=5(2)\)

Từ \((1);(2)\Rightarrow \sqrt{3(x+1)^2+4}+\sqrt{5(x+1)^2+9}\geq 5\geq 5-(x+1)^2\)

Dấu "=" xảy ra khi $(x+1)^2=0$ hay $x=-1$ (thỏa mãn)

Vậy pt có nghiệm $x=-1$

1)

ĐK: \(x\geq 5\)

PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)

2)

ĐK: \(x\geq -1\)

\(\sqrt{x+1}+\sqrt{x+6}=5\)

\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)

\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)

Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$

\(\Rightarrow x=3\) (thỏa mãn)

Vậy .............

3)

ĐK: \(x^2-6x+7\geq 0\)

Đặt \(\sqrt{x^2-6x+7}=a(a\geq 0)\) \(\Rightarrow x^2-6x=a^2-7\)

PT trở thành: \(a^2-7+a=5\Leftrightarrow a^2+a-12=0\)

\(\Leftrightarrow (a-3)(a+4)=0\Rightarrow a=3\) (do \(a\geq 0)\)

\(\Rightarrow \sqrt{x^2-6x+7}=3\)

\(\Rightarrow x^2-6x+7=9\)

\(\Leftrightarrow x^2-6x-2=0\) \(\Rightarrow x=3\pm \sqrt{11}\) (đều thỏa mãn)

a)  ĐK:  \(x\ge5\)

 \(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)

\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)

\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)

\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\) (t/m)

Vậy

b)  \(-5x+7\sqrt{x}=-12\)

\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)

\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)

đến đây tự làm

c) d) e) bạn bình phương lên

f)  \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)

             \(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)

           \(\ge\sqrt{9}+\sqrt{25}=8\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)

Vậy...