Đặt \(\sqrt{x^2+7x+8}=a\) thì ta có

\(a^2+a-20=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-5\left(l\right)\\a=4\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2+7x+8}=4\)

\(\Leftrightarrow x^2+7x-8=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}\)

\(x^2+7x+\sqrt{x^2+7x+8}=12\)

ĐK : \(x^2+7x+8\ge0\Leftrightarrow\orbr{\begin{cases}x\le\frac{-7-\sqrt{17}}{2}\\x\ge\frac{-7+\sqrt{17}}{2}\end{cases}}\)

Đặt \(t=x^2+7x\)

pt \(\Leftrightarrow t+\sqrt{t+8}=12\)

\(\Leftrightarrow\sqrt{t+8}=12-t\)( \(-8\le t\le12\))

Bình phương hai vế

\(\Leftrightarrow t+8=144-24t+t^2\)

\(\Leftrightarrow t^2-24t+144-t-8=0\)

\(\Leftrightarrow t^2-25t+136=0\)(*)

\(\Delta=b^2-4ac=\left(-25\right)^2-4\cdot136=625-544=81\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}t_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{25+\sqrt{81}}{2}=\frac{34}{2}=17\left(loai\right)\\t_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{25-\sqrt{81}}{2}=\frac{16}{2}=8\left(nhan\right)\end{cases}}\)

\(\Rightarrow x^2+7x=8\)

\(\Rightarrow x^2+7x-8=0\)

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

\(\Rightarrow x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-8\end{cases}\left(tm\right)}\)

Vậy phương trình có hai nghiệm \(\hept{\begin{cases}x_1=1\\x_2=-8\end{cases}}\)

\(ĐKXĐ:x^2+7x+8\ge0\Leftrightarrow\hept{\begin{cases}x\le\frac{-7-\sqrt{17}}{2}\\x\ge\frac{-7+\sqrt{17}}{2}\end{cases}}\)

Đặt \(x^2+7x=a\) nên

\(pt\Leftrightarrow a+\sqrt{a+8}=12\Leftrightarrow\sqrt{a+8}=12-a\)

\(\Leftrightarrow a+8=\left(12-a\right)^2=a^2-24a+144\)

\(\Leftrightarrow a^2-24a+144-a-8=0\)

\(\Leftrightarrow a^2-25a+136=0\)

\(\Leftrightarrow a^2-8a-17a+136=0\)

\(\Leftrightarrow a\left(a-8\right)-17\left(a-8\right)=0\)

\(\Leftrightarrow\left(a-17\right)\left(a-8\right)=0\Rightarrow\orbr{\begin{cases}a=17\\a=8\end{cases}}\)

Đến đây dễ rồi; lm

đk -3 =< x =< 10

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

\(\Leftrightarrow\dfrac{x+3-4}{\sqrt{x+3}+2}+\dfrac{10-x-9}{\sqrt{10-x}+3}=\left(x-6\right)\left(x-1\right)\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x+3}+2}+\dfrac{1-x}{\sqrt{10-x}+3}=\left(x-6\right)\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x+3}+2}-\dfrac{1}{\sqrt{10-x}+3}-x+6\ne0\right)=0\Leftrightarrow x=1\)(tm)

\(x^2-7x+8=2\sqrt{x}\)

\(\Leftrightarrow\left(x^2-6x+9\right)-x-1=2\sqrt{x}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{x}-4=0\left(1\right)\\x+\sqrt{x}-2=0\left(2\right)\end{cases}}\)

Giải (1): Ta đc: x= (9+căn17)/2

Giải (2) ta đc: x=1

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

Đặt \(a=\sqrt{x+5},b=\sqrt{x+2}\left(đk:a,b\ge0,a\ne b\right)\)

\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x^2+7x+10}\\a^2-b^2=x+5-x-2=3\end{matrix}\right.\)

PT trở thành: \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a=b\left(loại\right)\\a=1\\b=1\end{matrix}\right.\)

+ Với a=1

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

+ Với b=1

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

Vậy \(S=\left\{-1\right\}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a\\\sqrt{x+2=b}\end{matrix}\right.\)

Thì được:

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

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

Làm tiếp

\(ĐK:x\ge-2\)

\(PT\Leftrightarrow\dfrac{x+5-x-2}{\sqrt{x+5}+\sqrt{x+2}}\left(1+\sqrt{x^2+7x+10}\right)=3\\ \Leftrightarrow\dfrac{3\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)}{\sqrt{x+5}+\sqrt{x+2}}=3\\ \Leftrightarrow1+\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x+5}+\sqrt{x+2}\\ \Leftrightarrow\left(\sqrt{x+5}-1\right)\left(1-\sqrt{x+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1\\\sqrt{x+2}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x=-1\)

mình vừa lên lớp 9 , chưa học phương trình bậc 2 

a)2x³ + 7x² - x - 12 =0

=>2x³+x²-4x+6x²+3x-12=0

=>x(2x²+x-4)+3(2x²+x-4)=0

=>(x+3)(2x²+x-4)=0

=>x+3=0 hoặc 2x²+x-4=0

Xét x+3=0 <=>x=-3

Xét 2x²+x-4=0 ta dùng delta

\(\Delta=1^2-\left(-4\left(2.4\right)\right)=33>0\)

=>pt có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{33}}{4}\)

b)- x^3 + x^2 + 7x + 2 =0

=>-x³+3x²+x-2x²+6x+2=0

=>-x(x²-3x-1)+(-2)(x²-3x-1)=0

=>-(x+2)(x²-3x-1)=0

=>-(x+2)=0 hoặc x²-3x-1=0

Xét -(x+2)=0 <=>x=-2

Xét x²-3x-1=0 theo delta ta có:

\(\Delta=\left(-3\right)^2-\left(-4\left(1.1\right)\right)=13>0\)

=>pt cũng có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\)

xài hóc ne đi