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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a)
DK: x\(\ge\)-2,x\(\ge\)\(\dfrac{1}{2}\)
=>\(\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0\)
\(\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0\)
\(\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0\)
\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)
<=>25x+50=2x-1
=>23x=-51
=>x=\(-\dfrac{51}{23}\)(ko thỏa mãn dk)
=> phương trình vô nghiệm..
b)
ĐKXĐ:\(x\ge1,x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)(nhận)
Vậy S={1;8}
c) ĐKXĐ:
\(x\ge0\)
\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)
\(\Leftrightarrow-11\sqrt{2x}=-11\)
\(\Leftrightarrow\sqrt{2x}=1\)
\(\Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)
Câu a :\(\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0\) ( ĐK : \(x\ge\dfrac{1}{2}\) )
\(\Leftrightarrow\sqrt{4x+8}+\sqrt{9x+18}=\sqrt{2x-1}\)
\(\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=\sqrt{2x-1}\)
\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)
\(\Leftrightarrow25\left(x+2\right)=2x-1\)
\(\Leftrightarrow25x+50=2x-1\)
\(\Leftrightarrow23x=-51\)
\(\Leftrightarrow x=-\dfrac{51}{23}< -\dfrac{1}{2}\)
Vậy phương trình vô nghiệm .
Câu b :
\(\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0\) ( ĐK : \(x\ge1\) )
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-3\sqrt{\left(x-1\right)}=0\)
\(\Leftrightarrow\sqrt{\left(x-1\right)}\left(\sqrt{x+1}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
Vậy \(S=\left\{1;8\right\}\)
Câu c : \(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\) ( ĐK : \(x\ge\dfrac{5}{6}\) )
\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)
\(\Leftrightarrow-11\sqrt{2x}+11=0\)
\(\Leftrightarrow-11\left(\sqrt{2x}-1\right)=0\)
\(\Leftrightarrow\sqrt{2x}-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
Chúc bạn học tốt
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
Giải:
a) \(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}-3\sqrt{x-5.\dfrac{1}{9}}=\sqrt{1-x}\)
\(\Leftrightarrow2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
\(\Leftrightarrow\sqrt{x-5}=\sqrt{1-x}\)
\(\Leftrightarrow x-5=1-x\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
\(\Leftrightarrow\sqrt{4\left(x+2\right)}+2\sqrt{x+2}-\sqrt{9\left(x+2\right)}=1\)
\(\Leftrightarrow2\sqrt{x+2}+2\sqrt{x+2}-3\sqrt{x+2}=1\)
\(\Leftrightarrow\sqrt{x+2}=1\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\)
d) \(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}\right)^2-7^2}=2\)
\(\Leftrightarrow\sqrt{x-49}=2\)
\(\Leftrightarrow x-49=4\)
\(\Leftrightarrow x=53\)
Vậy ...
Câu c bạn xem lại đề, mình làm không ra, kết quả xấu
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
a) ⇔ |2x+3| = 8
⇒ \(\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x=5\\2x=-11\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy...
b) ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow3\sqrt{x}-7\sqrt{x}+6\sqrt{x}=8\)
\(\Leftrightarrow2\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\) (Vì \(x\ge0\) )
Vậy x = 16
c) ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{9\left(x-1\right)}=12\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
\(\Leftrightarrow x=17\)(TM)
Vậy x = 17