Bạn kiểm tra lại đề bài

Nếu bạn thiếu số 2 bên cạnh $\sqrt{2x^2+5x+3}$ thì có thể tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/tim-x-sao-cho-sqrt2x3sqrtx13x2sqrt2x25x3-16.235781793134

Đỗ Thanh Hải: uh ha, mình đã sửa lại rồi.

Bạn tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/giai-pt-sqrtx-2sqrt4-x2x2-5x-1.219493072549

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

ĐKXĐ: \(x^3-1\ge0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)\ge0\) 

mà \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)

\(2x^2+5x-1=7\sqrt{x^3-1}\Leftrightarrow2x^2+2x+2+3x-3=7\sqrt{x-1}\sqrt{x^2+x+1}\)

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{x-1}\sqrt{x^2+x+1}\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\\b=\sqrt{x^2+x+1}\end{matrix}\right.\left(a,b\ge0\right)\)

\(\Rightarrow\) pt trở thành \(2b^2+3a^2=7ab\Rightarrow2b^2-7ab+3a^2=0\)

\(\Rightarrow2b^2-6ab-ab+3a^2=0\Rightarrow2b\left(b-3a\right)-a\left(b-3a\right)=0\)

\(\Rightarrow\left(b-3a\right)\left(2b-a\right)=0\Rightarrow\left[{}\begin{matrix}b=3a\\2b=a\end{matrix}\right.\)

\(TH_1:b=3a\Rightarrow\sqrt{x^2+x+1}=3\sqrt{x-1}\)

\(\Rightarrow x^2+x+1=9\left(x-1\right)\Rightarrow x^2-8x+10=0\)

\(\Delta=\left(-8\right)^2-4.10=24\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{8-\sqrt{24}}{2}=4-\sqrt{6}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{8+\sqrt{24}}{2}=4+\sqrt{6}\end{matrix}\right.\)

\(TH_2:2b=a\Rightarrow2\sqrt{x^2+x+1}=\sqrt{x-1}\)

\(\Rightarrow4\left(x^2+x+1\right)=x-1\Rightarrow4x^2+3x+5=0\)

mà \(4x^2+3x+5=\left(2x\right)^2+2.2x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2+\dfrac{71}{16}=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{71}{16}>0\)

\(\Rightarrow\) loại 

Vậy pt có tập nghiệm \(S=\left\{4+\sqrt{6};4-\sqrt{6}\right\}\)

cái j zị

đề bị sao r đó

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow x-1=8x-3+2\sqrt{15x^2-13x+2}\)

\(\Leftrightarrow2-7x=2\sqrt{15x^2-13x+2}\)

Do \(x\le1\Rightarrow2-7x< 0\Rightarrow VP< 0\)

Mà \(VT>0\Rightarrow\) pt vô nghiệm

thanks

ĐK:\(\hept{\begin{cases}5x^2+27x+25\ge0\\x+1\ge0\\x^2-4\ge0\end{cases}}\)(*)

\(pt\Leftrightarrow\sqrt{5x^2+27x+25}=5\sqrt{x+1}+\sqrt{x^2-4}\)

\(\Leftrightarrow5x^2+27x+25=25x+25+x^2-4+10\sqrt{\left(x+1\right)\left(x^2-4\right)}\)

\(\Leftrightarrow4x^2+2x+4=10\sqrt{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow2x^2+x+2=5\sqrt{\left(x^2-x-2\right)\left(x+2\right)}\)

Đặt \(\hept{\begin{cases}\sqrt{x^2-x-2}=a\\\sqrt{x+2}=b\end{cases}}\)\(\Rightarrow2a^2+3b^2=5ab\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\2a=3b\end{cases}}\)..............

Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ 

1)\(\sqrt{4x^2+12x+9}=2-x\)

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

\(\Leftrightarrow\left|2x+3\right|=2-x\)

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

\(\)