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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
a/ \(\Rightarrow2x^2-3x-11=x^2-1\)
\(\Leftrightarrow x^2-3x-10=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Thay 2 nghiệm vào cả 2 căn thức thấy đều xác định
Vậy nghiệm của pt là ...
b/ \(\left\{{}\begin{matrix}x\ge-1\\2x^2+3x-5=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-6=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=2\\x=-3\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x=2\)
c/
\(\Leftrightarrow x^2+4x+4=3x^2-5x+14\)
\(\Leftrightarrow2x^2-9x+10=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{5}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\left\{{}\begin{matrix}-x-9\ge0\\\left(x-1\right)\left(2x-3\right)=\left(-x-9\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-9\\2x^2-5x+3=x^2+18x+81\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-9\\x^2-23x-78=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=26\left(ktm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
a)\(\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}=2\sqrt{\left(x+3\right)^2}\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}-2\sqrt{\left(x+3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+2}+\sqrt{x-1}-2\sqrt{x+3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x+2}+\sqrt{x-1}=2\sqrt{x+3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x+1+2\sqrt{\left(x-1\right)\left(x+2\right)}=4\left(x+3\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\2\sqrt{\left(x-1\right)\left(x+2\right)}=2x+11\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\4\left(x-1\right)\left(x+2\right)=4x^2+44x+121\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\-40x=129\end{cases}}\Rightarrow x=-3\) (thỏa)
b)\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)
Đk:\(x\ge-\frac{1}{3}\)
\(pt\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1-\left(\frac{3}{5}x+1\right)=\sqrt{3x+1}-\left(\frac{3}{5}x+1\right)\)
\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}-\frac{3}{5}x=\frac{3x+1-\left(\frac{3}{5}x+1\right)^2}{\sqrt{3x+1}+\frac{3}{5}x+1}\)
\(\Leftrightarrow\frac{3x\left(5-\sqrt{3x+10}\right)}{5\sqrt{3x+10}}=\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}\)
\(\Leftrightarrow\frac{3x\cdot\frac{25-3x-10}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)
\(\Leftrightarrow\frac{3x\cdot\frac{-3\left(x-5\right)}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)
\(\Leftrightarrow x\left(x-5\right)\left(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}\right)=0\)
Dễ thấy: \(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}< 0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Câu 1 :
Xét điều kiện:\(\hept{\begin{cases}x\ge5\\x\le1\end{cases}}\)(Vô lý)
Vậy pt vô nghiệm
Câu 2 :
\(2\sqrt{x+2}+2\sqrt{x+2}-3\sqrt{x+2}=1\)\(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\)
Vậy x=-1
Câu 3 :
\(\sqrt{3x^2-4x+3}=1-2x\)\(\Leftrightarrow3x^2-4x+3=1+4x^2-4x\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt{2}\)
Câu 4 :
\(4\sqrt{x+1}-3\sqrt{x+1}=4\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x=15\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
1) \(ĐK:\orbr{\begin{cases}0\le x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\end{cases}}\)
\(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\Leftrightarrow x-5+\sqrt{x^2-4x+1}=3\sqrt{x}-6\)\(\Leftrightarrow\frac{-6\left(x-4\right)}{x-5-\sqrt{x^2-4x+1}}=\frac{9\left(x-4\right)}{3\sqrt{x}+6}\Leftrightarrow\left(x-4\right)\left(\frac{9}{3\sqrt{x}+6}+\frac{6}{x-5-\sqrt{x^2-4x+1}}\right)=0\)
Xét phương trình \(\frac{9}{3\sqrt{x}+6}+\frac{6}{x-5-\sqrt{x^2-4x+1}}=0\Leftrightarrow\left(18\sqrt{x}-9\right)+9\left(x-\sqrt{x^2-4x+1}\right)=0\)\(\Leftrightarrow\frac{81\left(4x-1\right)}{18\sqrt{x}+9}+\frac{9\left(4x-1\right)}{x+\sqrt{x^2-4x+1}}=0\Leftrightarrow\left(4x-1\right)\left(\frac{81}{18\sqrt{x}+9}+\frac{9}{x+\sqrt{x^2-4x+1}}\right)=0\)
Dễ thấy \(\frac{81}{18\sqrt{x}+9}+\frac{9}{x+\sqrt{x^2-4x+1}}>0\)với mọi x thỏa mãn điều kiện nên 4x - 1 = 0 hay x = 1/4
Vậy phương trình có tập nghiệm S = {4; 1/4}
e làm câu dễ nhất ^^
\(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\left(đk:-1\le x\le4\right)\)
\(< =>\left(\sqrt{x+1}-1\right)+\left(\sqrt{4-x}-2\right)+\left(\sqrt{\left(x+1\right)\left(4-x\right)}-2\right)=0\)
\(< =>\frac{x}{\sqrt{x+1}+1}-\frac{x}{\sqrt{4-x}+2}+\frac{x\left(3-x\right)}{\sqrt{\left(x+1\right)\left(4-x\right)+2}}=0\)
\(< =>x=0\)
Điều kiện \(x\le\dfrac{10}{3}\)
Đặt \(\sqrt{10-3x}=p\left(p\ge0\right)\) thì pt đã cho \(\Rightarrow\left(\sqrt{4-3p}+1\right)\left(p+1\right)=9\) \(\left(p\le\dfrac{4}{3}\right)\)
Nếu \(\sqrt{4-3p}=h\left(h\ge0\right)\) thì \(\left(h+1\right)\left(p+1\right)=9\)
Đồng thời \(h^2+3p=\left(\sqrt{4-3p}\right)^2+3p=4-3p+3p=4\). Do vậy ta có hệ \(\left\{{}\begin{matrix}\left(h+1\right)\left(p+1\right)=9\\h^2+3p=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(h+1\right)\left(p+1\right)=9\\p=\dfrac{4-h^2}{3}\end{matrix}\right.\)
Từ đó \(\left(h+1\right)\left(\dfrac{4-h^2}{3}+1\right)=9\) \(\Leftrightarrow\left(h+1\right).\dfrac{7-h^2}{3}=9\) \(\Leftrightarrow\left(h+1\right)\left(7-h^2\right)=27\) \(\Leftrightarrow-h^3+7h+7-h^2=27\) \(\Leftrightarrow h^3+h^2-7h+20=0\) . Giải pt tìm được \(h=-4\) (loại do \(h\ge0\)). Vậy ta pt đã cho vô nghiệm.