gpt \(2x^2-5x-1=\sqrt{x+2}+\sqrt{4-x}\)
gpt :A= \(2x^2-5x-1=\sqrt{x+2}+\sqrt{4-x}\)
B= \(\sqrt{x^2-2x+5}+2\sqrt{4x+5}=x^3-2x^2+5x+4\)
Gpt: \(2x+2\sqrt{x^2+5x}+\sqrt{x+5}+\sqrt{x}=25\)
gpt:
\(x^4-2x^3+x=\sqrt{(x^2-x).2}\)
\(x(5x^3+2)-2(\sqrt{2x+1}-1)=0\)
\(\sqrt{x+\frac{3}{x}}=\frac{x^2-7}{2x+2}\)
c) Ta có:
\(\sqrt{x+\frac{3}{x}}=\frac{x^2+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{\sqrt{x^2+3}-2\sqrt{x}}{\sqrt{x}}=\frac{x^2-4x+3}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\sqrt{x^3+3x}+2x}=\frac{x^2-4x+3}{2\left(x+1\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\\sqrt{x^3+3x}+2x=2\left(x+1\right)\end{cases}}\)
+) \(x^2-4x+3=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
+) \(\sqrt{x^3+3x}+2x=2x+2\Rightarrow x=1\)
a/ Đặt \(\sqrt{2\left(x^2-x\right)}=a\)
\(\Rightarrow a^4-2a^2=a\)
\(\Leftrightarrow a\left(a+1\right)\left(a^2-a-1\right)=0\)
b/ \(x\left(5x^3+2\right)-2\left(\sqrt{2x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-1\right)^2+5x^4=0\)
\(\Leftrightarrow x=0\)
GPT : a, \(\sqrt{x^2-2x+5}\) + \(2\sqrt{4x+5}\)= \(x^3\)-\(2x^2\)+ \(5x+4\)
b,\(2x^2-5x-1=\sqrt{x-2}+\sqrt{4-x}\)
giúp mk vs : gpt :
A= \(\sqrt{x^2-2x+5}+2\sqrt{4x+5}=x^3-2x^2+5x+4\)
để mk làm cho ; bài này dùng liên hợp
pt<=> \(x+1-\sqrt{x^2-2x+5}+2x+4-2\sqrt{4x+5}+x^3-2x^2+2x-1=0\) ( ĐKXĐ: \(x\ge-\frac{5}{4}\))
<=> \(\frac{x^2+2x+1-\left(x^2-2x+5\right)}{x+1+\sqrt{x^2-2x+5}}+\frac{\left(2x+4\right)^2-4\left(4x+5\right)}{2x+4+2\sqrt{4x+5}}+\left(x-1\right)\left(x^2-x+1\right)=0\)
<=>: \(\frac{x^2+2x+1-x^2+2x-5}{x+1+\sqrt{x^2-2x+5}}+\frac{4x^2+16x+16-16x-20}{2x+4+2\sqrt{4x+5}}+\left(x-1\right)\left(x^2-x+1\right)=0\)
<=> \(\frac{4x-4}{x+1+\sqrt{x^2-2x+5}}+\frac{4x^2-4}{2x+4+2\sqrt{4x+5}}+\left(x-1\right)\left(x^2-x+1\right)=0\)
<=> \(\left(x-1\right)\left(\frac{4}{x+1+\sqrt{x^2-2x+5}}+\frac{4x+4}{2x+4+2\sqrt{4x+5}}+x^2-x+1\right)=0\)
<=> x=1 ( vì \(x\ge-\frac{5}{4}\)nên cái trong ngoặc thứ 2 khác 0)
vậy x=1
gpt:\(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(1-x\right)\left(x+3\right)\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-x^2-2x\)
\(\sqrt{x^2-x+2}+\sqrt{x^2-3x+6}=2x\)
GPT :
\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)
\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)
\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)
\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)
\(\Leftrightarrow\left(2x^3-8x^2+8x\right)+\left[\left(2x-3\right)-2\sqrt{2x-3}+1\right]=0\)
\(\Leftrightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)
Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)
Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.
x^3-4x^2+5x-1-căn 2x-3=0
=>\(x^3-4x^2+5x-2-\left(\sqrt{2x-3}-1\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2-\dfrac{2x-3-1}{\sqrt{2x-3}+1}=0\)
=>\(\left(x-2\right)\left[\left(x-1\right)\left(x-2\right)-\dfrac{2}{\sqrt{2x-3}+1}\right]=0\)
=>x-2=0
=>x=2
gpt a/ \(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)
b/ \(2\sqrt{1-x}-\sqrt{1+x}+3\sqrt{1-x^2}=3-x\)
b) Đặt \(u=\sqrt{1-x}\); \(v=\sqrt{1+x}\)
phương trình trở thành
\(2u-v+3uv=u^2+2\)\(\Rightarrow u^2-2u+v-3uv+2=0\)
lại có \(u^2+v^2=2\)
\(\Rightarrow u^2-2u-3uv+v+u^2+v^2=0\)
\(\Rightarrow\left(u-v-1\right)\left(2u-v\right)=0\)
đến đây thì easy rồi
a)
Đặt \(\sqrt{2x+1}=t\) ;\(\sqrt{x}=k\)
Phương trình trở thành
\(\left(3k^2+t^2\right)t-\left(3t^2+k^2\right)k-1=0\)
\(\Leftrightarrow3k^2t+t^3-3t^2k-k^3-1=0\)
\(\Leftrightarrow\left(t-k\right)\left(t^2+kt+k^2\right)-3tk\left(t-k\right)-1=0\)
\(\Leftrightarrow\left(t-k\right)^3-1=0\)
\(\Leftrightarrow\left(t-k-1\right)\left(\left(t-k\right)^2+t-k+1\right)=0\)
do t > k => t - k > 0
\(\Rightarrow\left(t-k\right)^2+t-k+1>0\)
\(\Rightarrow t-k-1=0\)
\(\Leftrightarrow t=1+k\)\(\Leftrightarrow\sqrt{2x+1}=1+\sqrt{x}\)
\(\Leftrightarrow2x+1=x+2\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
END
a)\(\left(5x+1\right)\sqrt{2x+1}-\left(7x+3\right)\sqrt{x}=1\)
ĐK:\(x\ge 0\)
\(\Leftrightarrow\left(5x+1\right)\sqrt{2x+1}-\left(\dfrac{31}{2}x+1\right)-\left(\left(7x+3\right)\sqrt{x}-\dfrac{31}{2}x\right)=0\)
\(\Leftrightarrow\dfrac{\left(5x+1\right)^2\left(2x+1\right)-\left(\dfrac{31}{2}x+1\right)^2}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{x\left(7x+3\right)^2-\left(\dfrac{31}{2}x\right)^2}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)
\(\Leftrightarrow\dfrac{\dfrac{1}{4}x\left(200x+19\right)\left(x-4\right)}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{\dfrac{1}{4}x\left(x-4\right)\left(196x-9\right)}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}=0\)
\(\Leftrightarrow\dfrac{1}{4}x\left(x-4\right)\left(\dfrac{200x+19}{\left(5x+1\right)\sqrt{2x+1}+\dfrac{31}{2}x-1}-\dfrac{196x-9}{\left(7x+3\right)\sqrt{x}+\dfrac{31}{2}x}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Nghe t đi phần nào khó cho qua :)) b tương tự
gpt \(\sqrt{5x^3-1}+\sqrt[3]{2x-1}+x-4=0\)