\(-x^2+3x+4\sqrt{\left(3-x\right)\left(x+1\right)}=10\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
A) \(\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}=2\sqrt{\left(x+3\right)^2}\)
B)\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)
a)\(\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}=2\sqrt{\left(x+3\right)^2}\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}-2\sqrt{\left(x+3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+2}+\sqrt{x-1}-2\sqrt{x+3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x+2}+\sqrt{x-1}=2\sqrt{x+3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x+1+2\sqrt{\left(x-1\right)\left(x+2\right)}=4\left(x+3\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\2\sqrt{\left(x-1\right)\left(x+2\right)}=2x+11\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\4\left(x-1\right)\left(x+2\right)=4x^2+44x+121\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\-40x=129\end{cases}}\Rightarrow x=-3\) (thỏa)
b)\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)
Đk:\(x\ge-\frac{1}{3}\)
\(pt\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1-\left(\frac{3}{5}x+1\right)=\sqrt{3x+1}-\left(\frac{3}{5}x+1\right)\)
\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}-\frac{3}{5}x=\frac{3x+1-\left(\frac{3}{5}x+1\right)^2}{\sqrt{3x+1}+\frac{3}{5}x+1}\)
\(\Leftrightarrow\frac{3x\left(5-\sqrt{3x+10}\right)}{5\sqrt{3x+10}}=\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}\)
\(\Leftrightarrow\frac{3x\cdot\frac{25-3x-10}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)
\(\Leftrightarrow\frac{3x\cdot\frac{-3\left(x-5\right)}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)
\(\Leftrightarrow x\left(x-5\right)\left(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}\right)=0\)
Dễ thấy: \(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}< 0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Giải pt:
a) \(x\left(x-4\right)\sqrt{-x^2+4x}+\left(x-2\right)^2=2\)
b) \(\left(x^2+1\right)^2=5-x\sqrt{2x^2+4}\)
c) \(2x^2+3x-14=2\sqrt[3]{2x^2+3x-10}\)
d) \(6x^2+2x+\sqrt[3]{3x^2+x+4}-10=0\)
e) \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
a/ ĐKXĐ: \(0\le x\le4\)
\(\left(x^2-4x\right)\sqrt{-x^2+4x}+x^2-4x+2=0\)
Đặt \(\sqrt{-x^2+4x}=a\ge0\)
\(-a^2.a-a^2+2=0\)
\(\Leftrightarrow a^3+a^2-2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+2a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a^2+2a+2=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-x^2+4x}=1\Leftrightarrow x^2-4x+1=0\Rightarrow...\)
b/ \(x^4+2x^2+x\sqrt{2x^2+4}-4=0\)
Đặt \(x\sqrt{2x^2+4}=a\Rightarrow x^2\left(2x^2+4\right)=a^2\Rightarrow x^4+2x^2=\frac{a^2}{2}\)
\(\frac{a^2}{2}+a-4=0\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=2\left(x>0\right)\\x\sqrt{2x^2+4}=-4\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^4+4x^2=4\\2x^4+4x^2=16\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=\sqrt{3}-1\\x^2=-\sqrt{3}-1\left(l\right)\\x^2=2\\x^2=-4\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)
c/ Đặt \(\sqrt[3]{2x^2+3x-10}=a\Rightarrow2x^2+3x=a^3+10\)
\(a^3+10-14=2a\)
\(\Leftrightarrow a^3-2a-4=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^2+2a+2\right)=0\Rightarrow a=2\)
\(\Rightarrow\sqrt[3]{2x^2+3x-10}=2\Rightarrow2x^2+3x-18=0\Rightarrow...\)
d/ \(\Leftrightarrow2\left(3x^2+x+4\right)+\sqrt[3]{3x^2+x+4}-18=0\)
Đặt \(\sqrt[3]{3x^2+x+4}=a\)
\(2a^3+a-18=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a^2+4a+9\right)=0\Rightarrow a=2\)
\(\Rightarrow\sqrt[3]{3x^2+x+4}=2\Rightarrow3x^2+x-4=0\Rightarrow...\)
e/ \(\Leftrightarrow x^2+5x+2-3\sqrt{x^2+5x+2}-2=0\)
Đặt \(\sqrt{x^2+5x+2}=a\ge0\)
\(a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=\frac{3+\sqrt{17}}{2}\\a=\frac{3-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+5x+2}=\frac{3+\sqrt{17}}{2}\Rightarrow x^2+5x-\frac{9+3\sqrt{17}}{2}=0\)
Bài cuối xấu quá, chắc nhầm số liệu
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
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1
\(\sqrt[3]{x^2-1}+\sqrt{3x^3-2}=3x-2\)
\(\sqrt{\left(2-x\right)\left(5-x\right)}=x+\sqrt{\left(2-x\right)\left(10-x\right)}\)
Bài 1:
Đk:\(1\le x\le2\)
\(pt\Leftrightarrow x^2-3x-10=-\sqrt{x^2-3x+2}\)
Đặt \(\sqrt{x^2-3x+2}=t\left(t\ge0\right)\) ta có:
\(t^2-12=-t\Leftrightarrow t^2+t-12=0\)\(\Leftrightarrow\left(t+4\right)\left(t-3\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-4\left(loai\right)\\t=3\end{array}\right.\)
Xét \(t=3\Leftrightarrow x^2-3x+2=3\)
\(\Leftrightarrow x^2-3x-1=0\)
\(\Delta=\left(-3\right)^2-\left[\left(-4\right).\left(1.1\right)\right]=13\)\(\Leftrightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\) (thỏa mãn)
Giải phương trình:
\(\frac{2\left(x-\sqrt{3}\right)\left(x-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
Giải phương trình :
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\) .
Dùng liên hợp.
pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)
\(-3\left(x-1\right)\left(x-\sqrt{3}\right)\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}\right)\)
\(+2\left(x-1\right)\left(x-\sqrt{2}\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+\sqrt{2}\right)=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)
\(=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(x+\sqrt{3}\right)\left(1-\sqrt{2}\right)\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left(x+1\right)\left(\sqrt{2}-\sqrt{3}\right)=3x-1\)
<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)
<=> \(x^2-3x+2=0\) phương trình bậc 2.
Em làm tiếp nhé!