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2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{\left(2\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{2}-\sqrt{3}\right)}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{\left(\sqrt{8}-\sqrt{3}\right)\left(\sqrt{8}+\sqrt{3}\right)}\)
\(=\sqrt{3}+1+\sqrt{3}-1+\frac{5\left(2\sqrt{2}+\sqrt{3}\right)}{5}-\frac{5\left(\sqrt{8}-\sqrt{3}\right)}{5}\)
\(=2\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{8}+\sqrt{3}\)
\(=4\sqrt{3}\)
Giải pt:
1/ \(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Rightarrow x=3\)
2/ \(\Leftrightarrow\sqrt{3}x^2=\sqrt{12}\Leftrightarrow x^2=\sqrt{4}=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
3/ \(\Leftrightarrow x-5=9\Rightarrow x=14\)
4/ Đề thiếu
5/ \(\Leftrightarrow\left|x-3\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
6/ \(\Leftrightarrow2\left|1-x\right|=6\)
\(\Leftrightarrow\left|1-x\right|=3\Leftrightarrow\left[{}\begin{matrix}1-x=3\\1-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
7/ \(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow x-1=49\Rightarrow x=50\)
8/ \(\Leftrightarrow x+1=2^3=8\)
\(\Rightarrow x=7\)
9/ \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)
10/ \(\Leftrightarrow\sqrt{2}x=\sqrt{50}\Leftrightarrow x=\sqrt{25}\Rightarrow x=5\)
11/ \(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
12/ \(\Leftrightarrow3-2x=\left(-2\right)^3=-8\)
\(\Leftrightarrow2x=11\Rightarrow x=\frac{11}{2}\)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)
\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)
\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)
\(=\frac{-29\sqrt{2}}{2}\)
2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)
\(=10-4\sqrt{3}+8+4\sqrt{3}\)
\(=18\)
3) ĐK để phương trình có nghiệm là: x≥0
Ta có: \(\sqrt{x^2-6x+9}=2x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{4x^2+1}=2x-1\)
\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow4x^2+1=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)
\(\Leftrightarrow4x=0\)
hay x=0(loại)
Vậy: S=∅
Giải pt :
1
a. ĐKXĐ : \(x\ge4\)
Ta có :
\(\sqrt{x+3}-\sqrt{x-4}=1\\ \Leftrightarrow\sqrt{x+3}=1+\sqrt{x-4}\\ \Leftrightarrow x+3=x-3+2\sqrt{x-4}\\ \Leftrightarrow6=2\sqrt{x-4}\)
\(\Leftrightarrow3=\sqrt{x-4}\\ \Leftrightarrow x-4=9\)
\(\Leftrightarrow x=13\) (TM ĐKXĐ)
Vậy \(S=\left\{13\right\}\)
b.ĐKXĐ : \(-3\le x\le10\)
Ta có :
\(\sqrt{10-x}+\sqrt{x+3}=5\\ \Leftrightarrow13+2\sqrt{-x^2+7x+30}=25\\ \Leftrightarrow\sqrt{-x^2+7x+30}=6\\ \Leftrightarrow-x^2+7x+30=36\\ \Leftrightarrow-x^2+7x-6=0\\ \Leftrightarrow-x^2+x+6x-6=0\\ \Leftrightarrow-x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(TMĐKXĐ\right)\\x=6\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy \(S=\left\{1;6\right\}\)
a) + b)
Câu hỏi của Trần Đức Mạnh - Toán lớp 9 | Học trực tuyến
c) Đk: \(-\sqrt{10}\le x\le\sqrt{10}\)
\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}=\left(x+3\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(N\right)\\\sqrt{10-x^2}=x-4\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow\left\{{}\begin{matrix}10-x^2=x^2-8x+16\\x\ge4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-8x+6=0\\x\ge4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=3\left(L\right)\\x=1\left(L\right)\end{matrix}\right.\\x\ge4\end{matrix}\right.\)
Kl: x=-3
Bạn làm phần a b mình Ko hiểu lắm. Mong bạn chỉ mình vài chỗ được ko. Nếu được thì cậu nt vs minh nhá