đặt \(\sqrt{2x^2+21x-11}=a\) và \(\sqrt{2x^2-9x+4}=b\)

==> \(a^2-b^2=30x-15\)

<=> \(\frac{a^2-b^2}{15}=2x-1\)

do đó pt đầu tên trở thành 

\(b+3\sqrt{\frac{a^2-b^2}{15}}=a\)

<=> \(\sqrt{\frac{a^2-b^2}{15}}=\frac{a-b}{3}\)

<=> \(\frac{a^2-b^2}{15}=\frac{a^2-2ab+b^2}{9}\)

<-=> \(9a^2-9b^2=15a^2-30ab+15b^2\)

<=> \(6a^2-30ab+24b^2=0\)

<=> \(a^2-5ab+4b^2=0\)

<=> \(\left(a-b\right)\left(a-4b\right)=0\)

<=> \(\orbr{\begin{cases}a=b\\a=4b\end{cases}}\)

đến đây bạn tự thay a;b vào rùi giải nốt nhé

\(ĐK:\left\{{}\begin{matrix}x\le\dfrac{1}{2};4\le x\\\dfrac{1}{2}\le x\\x\le-11;\dfrac{1}{2}\le x\end{matrix}\right.\Leftrightarrow x\le-11;4\le x\)

\(PT\Leftrightarrow\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\\ \Leftrightarrow\sqrt{2x-1}\left(\sqrt{x-4}-\sqrt{x+11}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\sqrt{x-4}-\sqrt{x+11}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x-4+x+11-2\sqrt{x^2+7x-44}=9\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\sqrt{x^2+7x-44}=2x-2\\ \Leftrightarrow\sqrt{x^2+7x-44}=x-1\\ \Leftrightarrow x^2+7x-44=x^2-2x+1\\ \Leftrightarrow9x=45\Leftrightarrow x=5\left(tm\right)\)

Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)

https://hoc24.vn/cau-hoi/giai-pt-sqrt2x2-9x43sqrt2x-1sqrt2x221x-11.2005877637936

làm r nha :vv

đk tự xử nha bạn

Đặt \(\sqrt{2x^2-9x+4}=a\)

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

\(\Rightarrow\sqrt{2x^2+21x-11}=\sqrt{a^2+15b^2}\) (\(a,b\ge0\))

PT \(\Rightarrow a-3b=\sqrt{a^2+15b^2}\) 

\(\Rightarrow a^2-6ab+9b^2=a^2+15b^2\)

\(\Leftrightarrow6b^2+6ab=0\)

\(\Leftrightarrow6b\left(a+b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=0\\a+b=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}\sqrt{2x-1}=0\\\sqrt{2x^2-9x+4}+\sqrt{2x-1}=0\end{cases}}\)

Bạn phải tìm đk của cả 2 pt mới trên nha!!!

PT 1 bạn tự tìm đk kết hợp với đk đầu bài rồi giải nha!

cái pt 2 nếu ko tìm được đk thích hợp thì ko cần giải nữa còn nếu tìm được thì bạn giải theo pp sau

Do \(\hept{\begin{cases}\sqrt{2x-1}\ge0\\\sqrt{2x^2-9x+4}\ge0\end{cases}}\)

\(\Rightarrow\sqrt{2x^2-9x+4}+\sqrt{2x-1}\ge0\)

Dấu = xảy ra khi \(\hept{\begin{cases}2x^2-9x+4=0\\2x-1=0\end{cases}\Rightarrow x=\frac{1}{2}}\)

Đây là mk giải tắt nha! bạn đối chiếu đk rồi loại nghiệm là ok rồi!

k mk nha!

\(DK:x\ge\frac{1}{2}\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-4\right)}-3\sqrt{2x-1}=\sqrt{\left(2x-1\right)\left(x+11\right)}\left(DK:x\ge4\right)\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{x-4}-3-\sqrt{x+11}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(1\right)\\\sqrt{x-4}-3-\sqrt{x+11}=0\left(2\right)\end{cases}}\)

PT(2)\(\Leftrightarrow\sqrt{x-4}=\sqrt{x+11}+3\)

\(\Leftrightarrow x-4=x+20+6\sqrt{x+11}\)

\(\Leftrightarrow-4=\sqrt{x+11}\) (Vo ly can cua mot bieu thuc khong the bang am)

Vay PT vo nghiem

2x² - 9x + 4 = 2x² - 8x - x + 4 = (2x -1).(x - 4)

2x² + 21x - 11 = 2x² + 22x - x - 11 = (2x -1).(x + 11)

Điều kiện: x \(\ge\) 4

PT <=> \(\sqrt{\left(2x-1\right)\left(x-4\right)}+3\sqrt{2x-1}=\sqrt{\left(2x-1\right)\left(x+11\right)}\)

<=> \(\sqrt{2x-1}\left(\sqrt{x-4}+3-\sqrt{x+11}\right)=0\)

<=> \(\sqrt{2x-1}=0\)  (1) hoặc \(\sqrt{x-4}-\sqrt{x+11}+3=0\) (2)

Giải (1) <=> x = 1/2 (Loại)

Giải (2) <=> \(\left(\sqrt{x-4}+3\right)^2=\left(\sqrt{x+11}\right)^2\)

<=> \(x-4+3+6\sqrt{x-4}=x+11\)

<=> \(\sqrt{x-4}=2\) <=> x = 8 (Thỏa mãn)

vậy x = 8

a, ĐKXĐ : Tự tìm hộ hen :)

Ta có : \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)

=> \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}-\sqrt{2x^2+21x-11}=0\)

=> \(\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\)

=> \(\sqrt{2x-1}\left(\sqrt{x-4}+3-\sqrt{x+11}\right)=0\)

=> \(\left[{}\begin{matrix}\sqrt{2x-1}=0\\\sqrt{x-4}+3=\sqrt{x+11}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-1=0\\x-4+6\sqrt{x-4}+9=x+11\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-1=0\\6\sqrt{x-4}=6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-1=0\\x-4=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=5\end{matrix}\right.\) ( TM )

Vậy ...

b, ĐKXĐ : Tiếp tục tìm hộ nha :)

Ta có : \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\frac{x-1}{x-2}}=3\)

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

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

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

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

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

=> \(\sqrt{1-x}=3\)

=> \(1-x=9\)

=> \(x=-8\left(TM\right)\)

Vậy ...

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-4\right)}+3\sqrt{2x-1}=\sqrt{\left(2x-1\right)\left(x+11\right)}\)

ĐK \(x\ge\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-4\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{x-4}+3-\sqrt{x+11}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(nh\right)\\\sqrt{x-4}+3=\sqrt{x+11}\end{matrix}\right.\)

\(\Leftrightarrow\left(\sqrt{x-4}+3\right)^2=x+11\)

\(\Leftrightarrow x+5+6\sqrt{x-4}=x+11\)

\(\Leftrightarrow x+5+6\sqrt{x-4}=x+11\)

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

\(\Leftrightarrow x=5\left(nh\right)\)

vậy \(S=\left\{\dfrac{1}{2};5\right\}\)

\(\sqrt{2x^2+21x-11}-3\sqrt{2x-1}=\sqrt{2x^2-9x+4}\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x+11\right)}-3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{x+11}-3-\sqrt{x-4}\right)=0\)

☘ Trường hợp 1:

\(\sqrt{2x-1}=0\)

⇔ 2x - 1 = 0

⇔ x = 0,5 (nhận)

☘ Trường hợp 1:

\(\sqrt{x+11}-3-\sqrt{x-4}=0\)

\(\Leftrightarrow\sqrt{x-4}+3=\sqrt{x+11}\)

\(\Leftrightarrow x-4+6\sqrt{x-4}+9=x+11\)

\(\Leftrightarrow6\sqrt{x-4}=6\)

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

⇔ x - 4 = 4

⇔ x = 5 (nhận)

☘ Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)