giải pt
\(\sqrt{x}-\sqrt{x-1}=\sqrt{x+8}-\sqrt{x+3}\)
Giải pt: \(\sqrt{x+2\sqrt{x+1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}+8\)
Kiểu dạng bài này là thường dưới căn cùng phép tính để đặt ẩn nên mình nghĩ là \(\sqrt{x+2\sqrt{x-1}}\) ...... mới đúng, còn nếu không phải thì bảo mình nhé và cách làm thì nó cũng giống cách mình làm thôi: )
ĐK: \(x\ge1\)
Đặt \(\sqrt{x-1}=t\left(t\ge0\right)\Rightarrow x=t^2+1\)
PT trở thành:
\(\sqrt{t^2+1+2t}+\sqrt{t^2+1-2t}=t+8\\ \Leftrightarrow\sqrt{\left(t+1\right)^2}+\sqrt{\left(t-1\right)^2}=t+8\\ \Leftrightarrow\left|t+1\right|+\left|t-1\right|=t+8\left(1\right)\)
Với \(0\le t< 1\) có:
(1) \(\Leftrightarrow t+1+1-t-t-8=0\)
\(\Leftrightarrow-6-t=0\\ \Leftrightarrow t=-6\left(loại\right)\)
Với \(t\ge1\) có:
(1) \(\Leftrightarrow t+1+t-1-t-8=0\)
\(\Leftrightarrow t-8=0\\ \Leftrightarrow t=8\left(nhận\right)\)
\(\Rightarrow x=t^2+1=8^2+1=64+1=65\)
Vậy nghiệm của PT là `x=65`
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
Giải PT :\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
NGUYỄN MINH TÀI Ok bí thì cx đừng gắt,t giải đoạn đó cho
\(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
\(VT=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\)
\(\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\("="\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x-1}\le3\Leftrightarrow4\le x-1\le9\)
\(\Leftrightarrow5\le x\le10\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)}^2+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Làm nốt nhé :v
giải pt :\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\) = 5
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)
Nếu \(\sqrt{x-1}\ge2\Rightarrow\left|\sqrt{x-1}-2\right|=\sqrt{x-1}-2\Rightarrow\sqrt{x-1}-2+\sqrt{x-1}+3=5\)
\(\Rightarrow2\sqrt{x-1}=4\Leftrightarrow x=5\)
Nếu \(0\le\sqrt{x-1}< 2\Rightarrow\left|\sqrt{x-1}-2\right|=2-\sqrt{x-1}\Rightarrow2-\sqrt{x-1}+\sqrt{x-1}+3=5\)
\(\Leftrightarrow2+3=5\)
giải pt :
a) \(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{x^4-1}\)
b0 \(4\sqrt{x+1}=x^2-5x+14\)
c) \(2x+3\sqrt{4-5x}+\sqrt{x+2}=8\)
d) \(\dfrac{x^2+x}{\sqrt{x^2+x+1}}=\dfrac{2-x}{\sqrt{x-1}}\)
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)
d.
ĐKXĐ: \(x>1\)
\(\Leftrightarrow\dfrac{x^2+x+1-1}{\sqrt{x^2+x+1}}=\dfrac{1-\left(x-1\right)}{\sqrt{x-1}}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2-1}{a}=\dfrac{1-b^2}{b}\)
\(\Leftrightarrow a-\dfrac{1}{a}=\dfrac{1}{b}-b\)
\(\Leftrightarrow a+b-\dfrac{a+b}{ab}=0\)
\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=0\)
\(\Leftrightarrow1-\dfrac{1}{ab}=0\)
\(\Leftrightarrow ab=1\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=1\)
\(\Leftrightarrow x^3-1=1\)
\(\Leftrightarrow x=\sqrt[3]{2}\)
Giải pt:
a,\(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
b,\(\sqrt{x-3+4\sqrt{x+1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\text{a) }\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\\ \Leftrightarrow\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\\ \Leftrightarrow\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+1}+\sqrt{\left(2x-1\right)-2\sqrt{2x-1}+1}=2\\ \Leftrightarrow\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)
Với \(x\ge1\Leftrightarrow\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\sqrt{2x-1}+1+\sqrt{2x-1}-1=2\\ \Leftrightarrow2\sqrt{2x-1}=2\\ \Leftrightarrow2x-1=1\\ \Leftrightarrow x=1\left(T/m\right)\)
Với \(x< 1\Leftrightarrow\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)
\(\Leftrightarrow0x=0\left(Nghiệm\text{ }đúng\text{ }\forall x\right)\\ \Leftrightarrow x< 1\)
Vậy pt có nghiệm \(x\le1\)
Giải pt :
a) \(x^2+3x\sqrt[3]{3x+3}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}\)
b) \(\sqrt{\left(x-1\right)\left(3-x\right)}+\sqrt{x+2}=\sqrt{x-1}+\sqrt{3-x}+\frac{x}{2}\)
giải pt :
a, \(\sqrt[3]{2-x}=1-\sqrt{x-1}\)
b, \(2\sqrt[3]{3x-2}+3\sqrt{6-5x}-8=0\)
c, \(\left(x+3\right)\sqrt{-x^2-8x+48}=x-24\)
d, \(\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{\left(7+x\right)\left(2-x\right)}=3\)
e, \(\dfrac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
giải pt: \(\sqrt{x+3}+\sqrt{1-x}=2-8\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\sqrt{x+3}+\sqrt{1-x}=2-8\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow\sqrt{x+3}+\sqrt{1-x}-2+8\sqrt{\left(x+3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x+3}-\frac{x+3}{\sqrt{1-x}+2}+8\sqrt{\left(x+3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(1-\frac{\sqrt{x+3}}{\sqrt{1-x}+2}+8\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\sqrt{x+3}=0\)
\(\Leftrightarrow x=-3\)