a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

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

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

\(\Leftrightarrow a=b\)

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

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

a: ĐKXĐ: \(x\ge\sqrt2\)

\(\sqrt{3x^2-7x+3}+\sqrt{x^2-3x+4}=\sqrt{3x^2-5x-1}+\sqrt{x^2-2}\)

=>\(\sqrt{3x^2-7x+3}-\sqrt{3x^2-5x+1}=\sqrt{x^2-2}-\sqrt{x^2-3x+4}\)

=>\(\frac{3x^2-7x+3-3x^2+5x-1}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}=\frac{x^2-2-x^2+3x-4}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

=>\(\frac{-2x+4}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}=\frac{3x-6}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

=>\(\frac{-2x+4}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}-\frac{3x-6}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}=0\)

=>\(\left(x-2\right)\left(\frac{-2}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}-\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\right)=0\)

=>x-2=0

=>x=2(nhận)

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

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

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

lamf nốt 4

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt[3]{x+6}-2+\sqrt{x-1}-1=x^2-4\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{x-2}{\sqrt[]{x-1}+1}=\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}=x+2\left(1\right)\end{matrix}\right.\)

Xét (1), do \(x\ge1\Rightarrow\left\{{}\begin{matrix}x+2\ge3\\\dfrac{1}{\sqrt[3]{\left(x+6\right)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt[]{x-1}+1}< \dfrac{1}{4}+\dfrac{1}{1}< 3\\\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)

em cảm ơn ạ

a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))

Vì hai vế ko âm, bp 2 vế ta được:

2x² - 3 = 4x - 3

\(\Leftrightarrow\) 2x² = 4x

\(\Leftrightarrow\) x² = 2x

\(\Leftrightarrow\) x² - 2x = 0

\(\Leftrightarrow\) x(x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

Vậy S = {2}

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

Vì hai vế ko âm, bp 2 vế ta được:

2x - 1 = x - 1

\(\Leftrightarrow\) x = 0 (KTM)

Vậy x = \(\varnothing\)

c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)

Vì hai vế ko âm, bp 2 vế ta được:

x² - x - 6 = x - 3

\(\Leftrightarrow\) x² - 2x - 3 = 0

\(\Leftrightarrow\) x² - 3x + x - 3 = 0

\(\Leftrightarrow\) x(x - 3) + (x - 3) = 0

\(\Leftrightarrow\) (x - 3)(x + 1) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy S = {3}

d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))

Vì hai vế ko âm, bp 2 vế ta được:

x² - x = 3x - 5

\(\Leftrightarrow\) x² - 4x + 5 = 0

\(\Leftrightarrow\) x² - 4x + 4 + 1 = 0

\(\Leftrightarrow\) (x - 2)² + 1 = 0

Vì (x - 2)² \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)² + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!

Nguyễn Lê Phước Thịnh nhờ anh xíu ạ

Không biết sao bạn cho thêm \(x\in Z\) vào cuối câu nhỉ? Giải pt nghiệm nguyên lai pt vô tỉ à :v

Bài làm :

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x-1}=b\\\sqrt{x+2}=c\end{matrix}\right.\)

\(pt\Leftrightarrow ac+ab+6=3a+2b+2c\)

\(\Leftrightarrow ac+ab+6-3a-2b-2c=0\)

\(\Leftrightarrow c\left(a-2\right)+b\left(a-2\right)-3\left(a-2\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(b+c-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\\b+c=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x-1}+\sqrt{x+2}=3\end{matrix}\right.\)

+) TH1: \(\sqrt{x+1}=2\)

\(\Leftrightarrow x+1=4\)

\(\Leftrightarrow x=3\) ( thỏa )

+) TH2: \(\sqrt{x-1}+\sqrt{x+2}=3\)

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

\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(x+2\right)}=8-2x\)

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

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

\(\Leftrightarrow x^2+x-2=x^2-8x+16\)

\(\Leftrightarrow9x=18\)

\(\Leftrightarrow x=2\) ( thỏa )

Vậy \(x\in\left\{2;3\right\}\).

ghê à nha, em tính liên hợp nhưng thôi, thấy anh làm r:)

a, ĐKXĐ: ...

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

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

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)

b: ĐKXĐ: \(\begin{cases}5x^2+14x+9\ge0\\ x+1\ge0\\ x^2-x-2\ge0\end{cases}\)

=>(5x+9)(x+1)>=0 và x>=-1 và (x-2)(x+1)>=0

=>(x>=-1 hoặc x<=-9/5) và x>=-1 và (x>=2 hoặc x<=-1)

=>x=-1 hoặc x>=2

Ta có: \(\sqrt{5x^2+14x+9}-5\sqrt{x+1}=\sqrt{x^2-x-2}\)

=>\(\sqrt{\left(5x+9\right)\left(x+1\right)}-5\sqrt{x+1}-\sqrt{\left(x-2\right)\left(x+1\right)}=0\)

=>\(\sqrt{x+1}\left(\sqrt{5x+9}-5-\sqrt{x-2}\right)=0\)

TH1: x+1=0

=>x=-1(nhận)

TH2: \(\sqrt{5x+9}-\sqrt{x-2}-5=0\)

=>\(\sqrt{5x+9}-\sqrt{x-2}=5\)

=>\(5x+9+x-2-2\cdot\sqrt{\left(5x+9\right)\left(x-2\right)}=25\)

=>\(2\cdot\sqrt{\left(5x+9\right)\left(x-2\right)}=6x-11-25=6x-36\)

=>\(\sqrt{\left(5x+9\right)\left(x-2\right)}=3x-18\)

=>\(\begin{cases}3x-18\ge0\\ \left(5x+9\right)\left(x-2\right)=\left(3x-18\right)^2\end{cases}\Rightarrow\begin{cases}x\ge6\\ 9x^2-108x+324=5x^2-10x+9x-18\end{cases}\)

=>\(\begin{cases}x\ge6\\ 9x^2-108x+324-5x^2+x+18=0\end{cases}\Rightarrow\begin{cases}x\ge6\\ 4x^2-107x+342=0\end{cases}\)

\(4x^2-107x+342=0\)

\(\Delta=\left(-107\right)^2-4\cdot4\cdot342=5977\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{107-\sqrt{5977}}{2\cdot4}=\frac{107-\sqrt{5977}}{8}\left(loại\right)\\ x=\frac{107+\sqrt{5977}}{2\cdot4}=\frac{107+\sqrt{5977}}{8}\left(nhận\right)\end{array}\right.\)