giải pt \(\sqrt{x-1}+\sqrt{2x+5}=8\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KS
1
28 tháng 2 2021
\(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\)
\(\Leftrightarrow11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{11x-5-2x^2}\)
\(\Leftrightarrow121\left(5-x\right)+176\sqrt{\left(5-x\right)\left(2x-1\right)}+64\left(2x-1\right)=576+144\sqrt{11x-5-2x^2}\)\(+9\left(11x-5-2x^2\right)\)
\(\Leftrightarrow605-121x+176\sqrt{11x-5-2x^2}+128x-64=576+144\sqrt{11x-5-2x^2}\)\(+99x-18x^2\)
\(\Leftrightarrow176\sqrt{11x-5-2x^2}-144\sqrt{11x-5-2x^2}=531+99x-18x^2-541-7x\)
\(\Leftrightarrow32\sqrt{11x-5-2x^2}=-10+92x-18x^2\)
\(\Leftrightarrow16\sqrt{11x-5-2x^2}=-5+46x-9x^2\)
\(\Leftrightarrow256\left(11x-5-2x^2\right)=25+2116x^2+81x^4-460x+90x^2-823x^3\)
\(\Leftrightarrow2816x-1280-512x^2=25+2206x^2+81x^4-460x-823x^3\)
\(\Leftrightarrow9\left(364x-145-302x^2-9x^4+92x^3\right)=0\)
\(\Leftrightarrow-9x^4+92x^3-302x^2+364x-145=0\)
\(\Leftrightarrow-\left(x-1\right)\left(9x^3-83x^2+219x-145\right)=0\)
\(\Leftrightarrow-\left(x-1\right)\left(x-1\right)\left(9x^2-74x+145\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\left(9x-29\right)\left(x-5\right)=0\Leftrightarrow\)x=1; x=29/9; x=5
\(\Leftrightarrow11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{11x-5-2x^2}\)
NQ
giải pt
a)\(\dfrac{1}{x+1}+\dfrac{3}{2x+1}=\dfrac{8}{x-2}\)
b)\(\sqrt{2x+1}+\sqrt{3-x}=\sqrt{3x+5}\)
0
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 8 2021
ĐKXĐ: \(x\ne-1\)
\(\dfrac{6x^2+4x+8}{x+1}=5\sqrt{2x^2+3}\)
\(\Rightarrow6x^2+4x+8=5\left(x+1\right)\sqrt{2x^2+3}\)
\(\Leftrightarrow2\left(2x^2+3\right)-5\left(x+1\right)\sqrt{2x^2+3}+2\left(x+1\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+3}=a\\x+1=b\end{matrix}\right.\)
\(\Rightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2+3}=2\left(x+1\right)\\2\sqrt{2x^2+3}=x+1\end{matrix}\right.\) (\(x\ge-1\))
\(\Rightarrow\left[{}\begin{matrix}2x^2+3=4\left(x+1\right)^2\\4\left(2x^2+3\right)=\left(x+1\right)^2\end{matrix}\right.\) (\(x\ge-1\))
\(\Leftrightarrow...\)
TN
8 tháng 9 2017
a)\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left|1-x\right|+\left|x-2\right|=3\)
Có: \(VT=\left|1-x\right|+\left|x-2\right|\)
\(\ge\left|1-x+x-2\right|=3=VP\)
Khi \(x=0;x=3\)
b)\(\sqrt{x^2-10x+25}=3-19x\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=3-19x\)
\(\Leftrightarrow\left|x-5\right|=3-19x\)
\(\Leftrightarrow x^2-10x+25=361x^2-114x+9\)
\(\Leftrightarrow-360x^2+104x+16=0\)
\(\Leftrightarrow-5\left(5x-2\right)\left(9x+1\right)=0\)
\(\Rightarrow x=\frac{2}{5};x=-\frac{1}{9}\)
c)\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
\(\Leftrightarrow2\sqrt{2x-3}+5=5\)\(\Leftrightarrow\sqrt{2x-3}=0\Leftrightarrow x=\frac{3}{2}\)
NH
0
DT
0
GD
24 tháng 11 2017
\(\sqrt{x-5}+\sqrt{x-3}-2\sqrt{x^2+2x-8}+4=0\left(1\right)\\ \Leftrightarrow\sqrt{x-5}+\sqrt{x-3}+4=2\sqrt{x^2+2x-8}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-5\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x\ge3\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-5}+\sqrt{x-3}+4=2\sqrt{x^2+2x-8}\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2+\left(\sqrt{x-3}\right)^2+4^2=\left(2\sqrt{x^2+2x-8}\right)^2\\ \Leftrightarrow x-5+x-3+16=4.\left(x^2+2x-8\right)\\ \Leftrightarrow x-5+x-3+16=4x^2+8x-32\\ \Leftrightarrow x-5+x-3+16-4x^2-8x+32=0\\ \Leftrightarrow-4x^2-6x+40=0\)
Ta có: \(\Delta=b^2-4ac=\left(-6\right)^2-4.\left(-4\right).40=676\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-6\right)+\sqrt{676}}{2.\left(-4\right)}=-4\left(nhận\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-6\right)-\sqrt{676}}{2.\left(-4\right)}=\dfrac{5}{2}=2,5\left(loại\right)\end{matrix}\right.\)
Vậy phương trình (1) không có nghiệm thỏa mãn.
GD
24 tháng 11 2017
Mình nhầm chỗ \(x_1=-4\) là loại mà mình nhấn nhầm là nhận!
Điều kiện \(x\ge1\)
Bình phương 2 vế ta được: \(x-1+2x+5+2\sqrt{\left(x-1\right)\left(2x+5\right)}=64\)
\(2\sqrt{\left(x-1\right)\left(2x+5\right)}=60-3x\). Bình phương 2 vế ta được:
\(4\left(x-1\right)\left(2x+5\right)=\left(60-3x\right)^2\)
<=> 4(2x2+3x-5)=9x2-360x+3600
<=> x2-372x+3620=0
<=> x2-10x-362x+3620=0
<=> x(x-10)-362(x-10)=0
<=> (x-10)(x-362)=0
<=> \(\hept{\begin{cases}x-10=0\\x-362=0\end{cases}}\)<=> \(\hept{\begin{cases}x_1=10\\x_2=362\end{cases}}\)