Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\left(3x+1\right)\sqrt{3x+1}=8x^2+5x+1\)
\(pt\Leftrightarrow\left(3x+1\right)\sqrt{3x+1}=8x^2+5x+1\)
\(\Leftrightarrow\frac{\left(3x+1\right)^3-1}{\left(3x+1\right)\sqrt{3x+1}+1}=8x^2+5x\)
\(\Leftrightarrow\frac{\left(3x+1-1\right)\left[\left(3x+1\right)^2+3x+2\right]}{\left(3x+1\right)\sqrt{3x+1}+1}=x\left(8x+5\right)\)
\(\Leftrightarrow\frac{9x\left(3x^2+3x+1\right)}{\left(3x+1\right)\sqrt{3x+1}+1}-x\left(8x+5\right)=0\)
\(\Leftrightarrow x\left(\frac{9\left(3x^2+3x+1\right)}{\left(3x+1\right)\sqrt{3x+1}+1}-\left(8x+5\right)\right)=0\)
\(\Rightarrow x=0\), nghiệm còn lại khó quá t gg =))
b)\(9x+17=6\sqrt{8x+1}+4\sqrt{x+3}\)
ĐK:\(x\ge-\frac{1}{8}\)
\(pt\Leftrightarrow9x-9=6\sqrt{8x+1}-18+4\sqrt{x+3}-8\)
\(\Leftrightarrow9\left(x-1\right)=\frac{36\left(8x+1\right)-324}{6\sqrt{8x+1}+18}+\frac{16\left(x+3\right)-64}{4\sqrt{x+3}+8}\)
\(\Leftrightarrow9\left(x-1\right)=\frac{288x-288}{6\sqrt{8x+1}+18}+\frac{16x-16}{4\sqrt{x+3}+8}\)
\(\Leftrightarrow9\left(x-1\right)-\frac{288\left(x-1\right)}{6\sqrt{8x+1}+18}-\frac{16\left(x-1\right)}{4\sqrt{x+3}+8}=0\)
\(\Leftrightarrow\left(x-1\right)\left(9-\frac{288}{6\sqrt{8x+1}+18}-\frac{16}{4\sqrt{x+3}+8}\right)=0\)
Suy ra x=1 là nghiệm duy nhất
\(x^2-5x+36=8\sqrt{3x+4}\)
\(\Leftrightarrow x^2-5x+36-8\sqrt{3x+4}=0\)
\(\Leftrightarrow\left(-8\sqrt{3x+4}+32\right)+\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow-8\left(\sqrt{3x+4}-4\right)+\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow-8.\frac{3x+4-16}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow-8.\frac{3x-12}{\sqrt{3x+4}+4}+\left(x-1\right)\left(x-4\right)=0\)
\(\left(x-4\right)\left(\frac{-24}{\sqrt{3x+4}+4}+x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\frac{-24}{\sqrt{3x+4}+4}+x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\-\frac{24}{\sqrt{3x+4}+4}+3+x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\-3.\frac{16-3x-4}{\left(\sqrt{3x+4}+4\right)^2}+\left(x-4\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(x-4\right)\left[\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1\right]=0\end{cases}}\)
Mà \(\frac{9}{\left(\sqrt{3x+4}+4\right)^2}+1>0\forall x\) nên \(x-4=0\Rightarrow x=4\)
Vật PT có nghiệm duy nhất là \(x=4\)
1. \(\sqrt{x^2+2x+3}=\sqrt{\left(x+1\right)^2+2}>0\)
=> Biểu thức luôn luôn có nghĩa với mọi x
2. \(\sqrt{x^2-2x+2}=\sqrt{\left(x-1\right)^2+1}>0\)
=> Biểu thức luôn luôn có nghĩa với mọi x
3. \(\sqrt{x^2+2x-3}=\sqrt{\left(x+1\right)^2-4}\)
\(\Rightarrow DK:\left(x+1\right)^2\ge4\)
4. \(\sqrt{2x^2+5x+3}=\sqrt{\left(\sqrt{2}x+\frac{5\sqrt{2}}{4}\right)^2-\frac{1}{8}}\)
\(\Rightarrow DK:\left(\sqrt{2}x+\frac{5\sqrt{2}}{4}\right)^2\ge\frac{1}{8}\)
K biết đúng k.. Sai thôi
1) tc : x2 + 2x +3 = x2 + 2x + 1 + 2 = (x+1)2 +2 > 0 vs mọi x
=> căn thức có nghĩa vs mọi x
2) tương tự câu 1: x2 - 2x + 2 = (x-1)2 +1 > 0 vs mọi x
=> căn thức có nghĩa vs mọi x
3) \(\sqrt{x^2+2x-3}\)có nghĩa <=> x2+2x-3\(\ge0\)
<=> (x+1)2 - 4 \(\ge0\)
<=> (x+1)2 \(\ge4\)
<=> x+1 \(\ge2\)
<=> x \(\ge1\)
4) \(\sqrt{2x^2+5x+3}\)có nghĩa <=> 2x2 +5x +3 \(\ge0\)
<=> 2x2 + 2x + 3x + 3 \(\ge0\)
<=> (2x+3)(x+1) \(\ge0\)
<=>\(\hept{\begin{cases}2x+3\ge0\\x+1\ge0\end{cases}}\) hoặc \(\hept{\begin{cases}2x+3\le0\\x+1\le0\end{cases}}\)
<=> \(\hept{\begin{cases}x\ge\frac{-3}{2}\\x\ge-1\end{cases}}\) hoặc \(\hept{\begin{cases}x\le\frac{-3}{2}\\x\le-1\end{cases}}\)
<=> \(\frac{-3}{2}\le x\le-1\)
Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
Bạn ghi lãi đề nhá!
\(x^2-5x+36=8\sqrt{3x+4}\)
tìm x