Đặt x^2+3x=a

=>\(a+2=3\sqrt{a}\)

=>a-3 căn a+2=0

=>(căn a-1)(căn a-2)=0

=>a=1 hoặc a=4

=>x^2+3x=1 hoặc x^2+3x=4

=>(x+4)(x-1)=0 và x^2+3x-1=0

=>\(x\in\left\{1;-4;\dfrac{-3+\sqrt{13}}{2};\dfrac{-3-\sqrt{13}}{2}\right\}\)

DỂ QUÁ!!!!!!!!!!!!!!!!!!!!!!!!

tui hk biết làm

em mới lớp 8 chuy ơi

E mới lớp 7 chị ơi

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

\(\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(1+\sqrt{x^2-x-2}\right)=3\left(DKXD:x\ge2\right)\)\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x-2}\right)\left(\sqrt{x+1}+\sqrt{x-2}\right)\left(1+\sqrt{x\left(x-2\right)+\left(x-2\right)}\right)=3\left(\sqrt{x+1}+\sqrt{x-2}\right)\)\(\Leftrightarrow\left\{\left(x+1\right)-\left(x-2\right)\right\}\left(1+\sqrt{\left(x+1\right)\left(x-2\right)}\right)=3\left(\sqrt{x+1}+\sqrt{x-2}\right)\)

\(\Leftrightarrow3\left(1+\sqrt{\left(x+1\right)\left(x-2\right)}\right)=3\left(\sqrt{x+1}+\sqrt{x-2}\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=1\\\sqrt{x-2}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\left(loai\right)\\x=3\left(nhan\right)\end{cases}}}\)

Vậy...

Đặt \(\hept{\begin{cases}\sqrt{x+1}=a\\\sqrt{x-2}=b\end{cases}}\left(a,b\ge0\right)\) thì ta có

\(\hept{\begin{cases}a^2-b^2=3\left(1\right)\\\left(a-b\right)\left(1+ab\right)=3\left(2\right)\end{cases}}\)

Lấy (1) - (2) vế theo vế ta được

\(a^2-b^2-\left(a-b\right)\left(1+ab\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-1-ab\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(1-a\right)\left(b-1\right)=0\)

Với a = b

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

\(\Leftrightarrow x+1=x-2\Leftrightarrow0x=3\left(l\right)\)

Với a = 1

\(\Leftrightarrow\sqrt{x+1}=1\Leftrightarrow x=0\left(l\right)\)

Với b = 1

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

Vậy PT có nghiệm là x = 3

\(x=-1\)Giao lưu thôi nhé

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

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

Đặt \(\hept{\begin{cases}\sqrt{x+5}=a\left(a\ge0\right)\\\sqrt{x+2}=b\left(b\ge0\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(ab+1\right)=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(ab+1-a-b\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2-b^2=3\\\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\end{cases}}\)

Với a = b thì

\(\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow0x=3\left(l\right)\)

Với a = 1 thì

\(\sqrt{x+5}=1\Leftrightarrow x=-4\left(l\right)\)

Với b = 1 thì

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

1.

đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)

có \(a^2+b^2=4\)

pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)

\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)

\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)

vì a,b>o nên \(a-b=\sqrt{2}\)

\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)

Bình phương 2 vế:

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

\(\Leftrightarrow\sqrt{4-x}=1\)

\(\Rightarrow x=3\)

Nếu đúng thì tích giùm mình cái nha!!!!!!!!!!!

2.ĐKXĐ D=R
Đặt \(a=\sqrt[3]{7-x},b=\sqrt[3]{x-5}\)
ta có: \(\hept{\begin{cases}a^3+b^3=2\\a^3-b^3=12-2x=2\left(6-x\right)\end{cases}}\)
Vậy ta có:

\(\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\Leftrightarrow\left(a-b\right)\left(2-\left(a+b\right)\left(a^2+ab+b^2\right)\right)=0\)
Th1: \(a-b=0\Leftrightarrow\sqrt[3]{7-x}=\sqrt[3]{x-5}\Leftrightarrow x=6\)
Th2: \(\hept{\begin{cases}\left(a+b\right)\left(a^2+ab+b^2\right)=2\\a^3+b^3=12\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)\left(a^2+ab+b^2\right)=2\\\left(a+b\right)\left(a^2-ab+b^2\right)12\end{cases}}\)
Từ đó suy ra: 

\(\frac{a^2-ab+b^2}{a^2+ab+b^2}=6\Leftrightarrow5a^2-7ab+6b^2=0\)
nếu \(b=0\Leftrightarrow\sqrt[3]{x-5}=0\Leftrightarrow x=5\)thay vào phương trình ta thấy không thỏa mãn.
nếu \(b\ne0\Rightarrow5a^2-7ab+5b^2=0\Leftrightarrow5\left(\frac{a}{b}\right)^2-7\frac{a}{b}+5=0\)(1)
phương trình (1) vô nghiệm với ẩn \(\frac{a}{b}\). nên trường hợp này không xảy ra.
vậy phương trình có duy nhất nghiệm x = 6.

Đk:  \(\hept{\begin{cases}x\ge2\\2x+3+\sqrt{x+2}\ge0\\2x+2-\sqrt{x+2}\ge0\end{cases}}\)

Đặt \(\sqrt{x+2}=t\left(t\ge0\right)\Rightarrow x=t^2-2\)

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

\(\Leftrightarrow4t^2-3+2\sqrt{\left(2t^2+t-1\right)\left(2t^2-t-2\right)}=4t^2+4t+1\)

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

\(\Leftrightarrow4t^4-11t^2-9t-2=0\)

\(\Leftrightarrow\left(2t+1\right)^2\left(t-2\right)\left(t+1\right)=0\)

Do \(t\ge0\) nên t = 2. Vậy \(\sqrt{x+2}=2\Rightarrow x=2\left(tm\right)\)

Vậy pt có nghiệm x = 2.

Chúc em học tốt!