Giải PT:
\(\sqrt{x}+\sqrt{x+8}=4\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)
\(\Rightarrow a^2+3-4a=0\)
=> (a - 3).(a - 1) = 0
=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)
Bình phương lên giải tiếp nhé!
c) Tương tư câu b nhé
ĐKXĐ: \(\left\{{}\begin{matrix}x+8>=0\\6-2x>=0\end{matrix}\right.\Leftrightarrow-8< =x< =3\)
\(PT\Leftrightarrow1=6\sqrt{x+8}-18+x+4\sqrt{6-2x}-8\)
\(\Leftrightarrow6\cdot\dfrac{x+8-9}{\sqrt{x+8}+3}+x-1+4\cdot\dfrac{6-2x-4}{\sqrt{6-2x}+2}=0\)
=>\(\left(x-1\right)\left(\dfrac{6}{\sqrt{x+8}+3}+1-\dfrac{4}{\sqrt{6-2x}+2}\right)=0\)
=>x-1=0
=>x=1
nhầm đề : \(\sqrt[4]{x+8}+\sqrt{x+4}=\sqrt{2x+3}+\sqrt{3x}\)
\(\sqrt[4]{x+8}+\sqrt{x+4}=\sqrt{2x+3}+\sqrt{3x}\)
\(\Leftrightarrow\sqrt[4]{x+8}-\sqrt{3}+\sqrt{x+4}-\sqrt{5}=\sqrt{2x+3}-\sqrt{5}+\sqrt{3x}-\sqrt{3}\)
\(\Leftrightarrow\frac{x+8-9}{\sqrt[4]{x+8}^3+\sqrt[4]{x+8}^2\sqrt{3}+3\sqrt[4]{x+8}+\sqrt{3}^3}+\frac{x+4-5}{\sqrt{x+4}+\sqrt{5}}=\frac{2x+3-5}{\sqrt{2x+3}+\sqrt{5}}+\frac{3x-3}{\sqrt{3x}+\sqrt{3}}\)
\(\Leftrightarrow\frac{x-1}{\sqrt[4]{x+8}^3+\sqrt[4]{x+8}^2\sqrt{3}+3\sqrt[4]{x+8}+\sqrt{3}^3}+\frac{x-1}{\sqrt{x+4}+\sqrt{5}}-\frac{2\left(x-1\right)}{\sqrt{2x+3}+\sqrt{5}}-\frac{3\left(x-1\right)}{\sqrt{3x}+\sqrt{3}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt[4]{x+8}^3+\sqrt[4]{x+8}^2\sqrt{3}+3\sqrt[4]{x+8}+\sqrt{3}^3}+\frac{1}{\sqrt{x+4}+\sqrt{5}}-\frac{2}{\sqrt{2x+3}+\sqrt{5}}-\frac{31}{\sqrt{3x}+\sqrt{3}}\right)=0\)
Dễ thấy : pt trong ngoặc vô nghiệm
\(\Rightarrow x-1=0\Rightarrow x=1\)
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
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\) = 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 \(0\le\sqrt{x-1}< 2\Rightarrow\left|\sqrt{x-1}-2\right|=2-\sqrt{x-1}\Rightarrow2-\sqrt{x-1}+\sqrt{x-1}+3=5\)
\(\Leftrightarrow2+3=5\)
Điều kiện xác định : \(x\ge2\)
Ta có : \(\sqrt{x+8+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{\left(x+7\right)-\sqrt{x+7}-6}=4\)
\(\Leftrightarrow\sqrt{x+7}+\sqrt{\left(x+7\right)-\sqrt{x+7}-6}-3=0\)
Đặt \(t=\sqrt{x+7},t\ge0\) , pt trở thành \(t+\sqrt{t^2-t-6}-3=0\)
\(\Leftrightarrow\left(t-3\right)+\sqrt{\left(t-3\right)\left(t+2\right)}=0\)
\(\Leftrightarrow\sqrt{t-3}\left(\sqrt{t-3}+\sqrt{t+2}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{t-3}=0\\\sqrt{t-3}+\sqrt{t+2}=0\end{array}\right.\)
Vì \(\sqrt{t-3}\ge0,\sqrt{t+2}\ge0\Rightarrow\sqrt{t-3}+\sqrt{t+2}\ge0\) . Dấu "=" không đồng thời xảy ra nên pt vô nghiệm.
Vậy t = 3 => x = 2
pt có nghiệm x = 2
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 \(x=-1\)
Bài này có nhiều cách đây là 2 cách mình có
C1:
Đặt a=can(x); b=can(x+8) (a;b>0)
ta có hệ pt
\(\hept{\begin{cases}a+b=4\\a^2-b^2=-8\end{cases}}\)
giải ra ta được
\(\hept{\begin{cases}a=1\\b=3\end{cases}< =>x=1}\)
C2: xét khoảng
Xét x=1 (thỏa mãn)
Xét x>1
\(\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x+8}>3\end{cases}}\)
suy ra VT>4>=VP (vô lí trái giả thiết loại)
Xét 0<x<1
\(\hept{\begin{cases}\sqrt{x}< 1\\\sqrt{x+8}< 3\end{cases}}\)
suy ra VT<4<=VP(vô lí trái giả thiết loại)
\(\sqrt{x}+\sqrt{x+8}=4\)
\(\Leftrightarrow\sqrt{x}+\frac{x+8}{\sqrt{x+8}}=4\)
\(\Leftrightarrow\frac{\sqrt{x}\left(\sqrt{x+8}\right)}{\sqrt{x+8}}+\frac{x+8}{\sqrt{x+8}}=4\)
\(\Leftrightarrow\frac{\sqrt{x^2+8x}}{\sqrt{x+8}}+\frac{x+8}{\sqrt{x+8}}=4\)
\(\Leftrightarrow\frac{\sqrt{x^2+8x}+x+8}{\sqrt{x+8}}=4\)
\(\Leftrightarrow\frac{\sqrt{x\left(x+8\right)}+\left(x+8\right)}{\sqrt{x+8}}=4\)
\(\Leftrightarrow\frac{\sqrt{x}.\sqrt{x+8}+\left(x+8\right)}{\sqrt{x+8}}=4\)
\(\Leftrightarrow\sqrt{x}+x+8=4\)
\(\Leftrightarrow\sqrt{x}+x+8-4=0\)
\(\Leftrightarrow\sqrt{x}+x+4=0\)
\(\Leftrightarrow\sqrt{x}=-4-x\)
\(\Leftrightarrow\hept{\begin{cases}-x-4\ge0\\x=\left(-x-4\right)^2\end{cases}}\Rightarrow\hept{\begin{cases}x\le4\\x=x^2-8x+16\left(\cdot\right)\end{cases}}\)
giai phuong trinh ( . )
\(\Leftrightarrow x=x^2-8x+16\)
\(\Leftrightarrow x^2-8x-x+16=0\)
\(\Leftrightarrow x^2-9x+16=0\)
ban tu lam tiep