Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3x-1\\t=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+2x-3}{-2}=1-x\\t=\dfrac{1-2x+3}{-2}=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{8}{3}\)

\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)

Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)

\(\left(t^2-6\right)^2-6-t=0\)

\(\Leftrightarrow t^4-12t^2-t+30=0\)

\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13,đkxđ:-1\le x\le7,\Leftrightarrow\left(\sqrt{7-x}+\sqrt{x+1}\right)^2=\left(x^2-6x+13\right)^2\Leftrightarrow7-x+x+1+2\sqrt{\left(7-x\right)\left(x+1\right)}=\left(x^2-6x+13\right)\left(x^2-6x+13\right)\Leftrightarrow8+2\sqrt{7x+8-x^2-x}=x^4-6x^3+13x^2-6x^3+36x^2-78x+13x^2-78x+169\Leftrightarrow8+2\sqrt{-x^2+6x+8}=x^4-12x^3+62x^2-120x+169\Leftrightarrow Bírồi:< \)

\(Chot=7-x\Rightarrow x=7-t\Rightarrow\sqrt{7-x}=\sqrt{7-7+t}=\sqrt{t}và\sqrt{x+1}=\sqrt{7-t+1}=\sqrt{8-t}vàx^2-6x+13=\left(7-t\right)^2-6\left(7-t\right)+13,tacópt:\sqrt{t}+\sqrt{8-t}=49-14t+t^2-42+6t+13\Leftrightarrow\sqrt{t}+\sqrt{8-t}=t^2-8t+20=t^2-2.4.t+16+4=\left(t-4\right)^2+4\Leftrightarrow\left(\sqrt{t}+\sqrt{8-t}\right)^2=\left[\left(t-4\right)^2+4\right]^2\Leftrightarrow t-t+8+2\sqrt{8t-t^2}=...\left(bítiếp\right)\)

\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13\left(đk:-1\le x\le7\right)\)

Với a,b>0 ta AD BĐT: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\) (tự CM nha ).Dấu "=" xảy ra<=>a=b (1)

AD bđt (1) có:

\(\sqrt{7-x}+\sqrt{x+1}\le2\sqrt{\frac{7-x+x+1}{2}}\)

\(\le2\sqrt{4}\) =4 (*)

Có x²-6x+13=(x-3)²+4 \(\ge4\) (**)

Từ (*),(**) => Dấu bằng xảy ra \(< =>\left\{{}\begin{matrix}7-x=x+1\\x-3=0\end{matrix}\right.\) \(< =>\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)\(< =>x=3\)(tm điều kiện của x)

Vậy x=3

đơn giản như đan rổ

1. đk: pt luôn xác định với mọi x

\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)

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

\(\Leftrightarrow\left|x-1\right|-\left|x-3\right|=10\)

Bạn mở dấu giá trị tuyệt đối như lớp 7 là ok rồi!

2.  đk: \(x\geq 1\)

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

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

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

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-3\sqrt{x-1}+5=0\)

Đến đây thì ổn rồi! bạn cứ xét khoảng rồi mở trị và bình phương 1 chút là ok cái bài!

a:

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

=>|x-3|=3

=>x-3=3 hoặc x-3=-3

=>x=0 hoặc x=6

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

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

=>\(\left|\sqrt{x-1}+1\right|=2\)

=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)

=>x-1=1

=>x=2

c:

ĐKXĐ: x>4/5

PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

d: ĐKXĐ: x-4>=0 và x+1>=0

=>x>=4

PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)

=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)

=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)

=>\(\sqrt{x^2-3x-4}=14-x\)

=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196

=>x<=14 và -3x-4=-28x+196

=>x<=14 và 25x=200

=>x=8(nhận)

a) \(\sqrt{x^2-6x+9}=3\)

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

\(\Leftrightarrow\left|x-3\right|=3 \)

TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)

Pt trở thành:

\(x-3=3\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=3+3\)

\(\Leftrightarrow x=6\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)

Pt trở thành:

\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=-3+3\)

\(\Leftrightarrow x=0\left(tm\right)\)

b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))

\(\Leftrightarrow x+2\sqrt{x-1}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4-x\)

\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)

\(\Leftrightarrow4x-4=16-8x+x^2\)

\(\Leftrightarrow x^2-12x+20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))

\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)

\(\Leftrightarrow5x-4=4x+8\)

\(\Leftrightarrow x=12\left(tm\right)\)