a) \(\sqrt{x+1}\).(x + 4) = (18 + x)\(\sqrt{6+x}\)- 3x - 40
b) 3 (\(\sqrt{x+9}\)- \(\sqrt{x+1}\)) = 4 - 4x
GIẢI PT HỘ MIK VS NHA <3 PT NÀO CX ĐC, CẢ 2 THÌ CÀNG TỐT ~ ^.^ ~
giải pt:
a) \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\)
b) \(3x+\sqrt{4x^2-8x+4}=1\)
c) \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)
giúp mk vs ạ mk cần gấp
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
Ai dậy r giúp vs :33 1 câu cx đc nhé :v toàn giải pt hết nhé
1) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}.\)
2) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}\)
3) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
4) \(\left(x+17\right)\sqrt{4-x}+\left(20-x\right)\sqrt{x+1}-9\sqrt{4-x}.\sqrt{x+1}=36\)
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
câu 2 ĐK \(x\ge1\)
\(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}=0\)
<=> \(\left(5x+8\right)\left(\sqrt{2x-1}-1\right)+7x\left(\sqrt{x+3}-2\right)+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)
<=>\(\left(5x+8\right).\frac{2x-2}{\sqrt{2x-1}+1}+7x.\frac{x+3-4}{\sqrt{x+3}+2}+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)
<=> \(\sqrt{x-1}\left(\frac{2\left(5x+8\right)\sqrt{x-1}}{\sqrt{2x-1}+1}+\frac{7x\sqrt{x-1}}{\sqrt{x+3}+2}+\left(x+26\right)+10\sqrt{x-1}\right)=0\)
Với \(x\ge1\)thì cái trong ngoặc >0
=> \(x=1\)
Vậy x=1
giải pt
a) \(\sqrt[3]{x+6}+\sqrt{x-1}=x^2-1\)
b) \(\sqrt[3]{x-9}+2x^2+3x=\sqrt{5x-1}+1\)
c) \(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
d) \(\sqrt{x+1}-2\sqrt{4-x}=\frac{5\left(x-3\right)}{\sqrt{2x^2+18}}\)
e) \(x^3+5x^2+6x=\left(x+2\right)\left(\sqrt{2x+2}+\sqrt{5-x}\right)\)
giải pt :
a, \(\sqrt{3x^2-7x+3}+\sqrt{x^2-3x+4}=\sqrt{3x^2-5x-1}+\sqrt{x^2-2}\)
b, \(\sqrt{x}+\sqrt{3-x}=x^2-x-2\)
c, \(\sqrt{x+6}+\sqrt{x-1}=x^2-1\)
ai có thể giúp mình giải bài này vs đc không mình đang cần rất gấp (làm chi tiết hộ mình nhé, xin cảm ơn)
Bài 4:
a, \(\sqrt{3x+4}-\sqrt{2x+1}=\sqrt{x+3}\)
b, \(\sqrt{2x-5}+\sqrt{x+2}=\sqrt{2x+1}\)
c, \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
d,\(\sqrt{x+9}=5-\sqrt{2x+4}\)
Bài 5:
a, \(\sqrt{x+4\sqrt{x}+4}=5x+2\)
b, \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)
c, \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
d,\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
Ví Dụ 1:
a, \(\sqrt{2x-1}=\sqrt{2}-1\)
b, \(\sqrt{x+5}=3-\sqrt{2}\)
c, \(\sqrt{3x^2}-\sqrt{12}=0\)
d, \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
Giair phương trình:
1) \(\sqrt[5]{32-x^2}-\sqrt[5]{1-x^2}=4\)
2) \(\sqrt{x}+\sqrt[4]{20-x}=4\)
3) \(x^3+1=2\sqrt{3x-1}\)
4) \(\sqrt[3]{x-1}+3=\sqrt[4]{82-x}\)
5)
\(a.\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(b.\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
giải pt :
a, \(3x^2+3x+2=\left(x+6\right)\sqrt{x^2-2x-3}\)
b, \(\sqrt{x}+\sqrt{x+1}=\sqrt{x^2+x}+1\)
c, \(\sqrt{x^2-8x+15}+\sqrt{x^2+2x-15}=\sqrt{x^2-9x+18}\)
c.
ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)
\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)
- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:
\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)
Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm
- Với \(x\le-5\) pt tương đương:
\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)
Do \(3-x>0\) pt trở thành:
\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)
\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)
\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))
\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)
\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)
a.
Kiểm tra lại đề, pt này không giải được
b.
ĐKXĐ: \(x\ge0\)
\(\sqrt{x\left(x+1\right)}-\sqrt{x}+1-\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}-1\right)-\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :
a, \(x^2+5x+2=4\sqrt{x^3+3x^2+x-1}\)
b, \(\sqrt{x+1}+x+3=\sqrt{1-x}+3\sqrt{1-x^2}\)
c,\(\left(2x-3\right)\sqrt{3+x}+2x\sqrt{3-x}=6x-8+\sqrt{9-x^2}\)
a, ĐK: \(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)
\(x^2+5x+2=4\sqrt{x^3+3x^2+x-1}\)
\(\Leftrightarrow x^2+2x-1+3\left(x+1\right)-4\sqrt{\left(x+1\right)\left(x^2+2x-1\right)}=0\)
TH1: \(x\ge-1\)
\(pt\Leftrightarrow\left(\sqrt{x^2+2x-1}-\sqrt{x+1}\right)\left(\sqrt{x^2+2x-1}-3\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=\sqrt{x+1}\\\sqrt{x^2+2x-1}=3\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
TH2: \(x< -1\)
\(pt\Leftrightarrow\left(\sqrt{-x^2-2x+1}-\sqrt{-x-1}\right)\left(\sqrt{-x^2-2x+1}-3\sqrt{-x-1}\right)=0\)
\(\Leftrightarrow...\)
Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi