\(\sqrt{x^6+37x^4+400x^2+1344}=\left(x^2+5\right)\left(3x^2-6-\sqrt{x^2+12}\right)+42\)
Giải phương trình !
Giải phương trình: \(\frac{\left(x^6+3x^4\sqrt{x^2-x+1}\right)\left(3+x-x^2\right)}{4\left(2+\sqrt{x^2-x+1}\right)\left(x^2-x+1\right)}=\sqrt{x^2-x+1}\left(2-\sqrt{x^2-x+1}\right)\)
Câu hỏi của Phương Boice - Toán lớp 8 - Học toán với OnlineMath
Giải phương trình: \(\frac{\left(x^6+3x^4\sqrt{x^2-x+1}\right)\left(3+x-x^2\right)}{4\left(2+\sqrt{x^2-x+1}\right)\left(x^2-x+1\right)}=\sqrt{x^2-x+1}\left(2-\sqrt{x^2-x+1}\right)\)
Đặt \(\sqrt{x^2-x+1}=a\left(ĐK:a>0\right)\)
\(pt\Leftrightarrow\frac{\left(x^6+3x^4a\right)\left(4-a^2\right)}{4\left(2+a\right)a^2}=a\left(2-a\right)\)
\(\Leftrightarrow\left(x^6+3x^4a\right)\left(4-a^2\right)=4a^3\left(4-a^2\right)\)
\(\Leftrightarrow\left(4-a^2\right)\left(x^6+3x^4a-4a^3\right)=0\)
TH1: \(4-a^2=0\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=2\left(n\right)\end{cases}}\)
Với a = 2 , \(\sqrt{x^2-x+1}=2\Rightarrow x^2-x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}+1}{2}\\x=\frac{-\sqrt{13}+1}{2}\end{cases}}\)
TH2: \(x^6+3x^4a-4a^3=0\Rightarrow x^6-x^4a+4x^4a-4x^2a^2+4x^2a^2-4a^3=0\)
\(\Leftrightarrow\left(x^2-a\right)\left(x^4+4x^2a+4a^2\right)=0\Leftrightarrow\left(x^2-a\right)\left(x^2+2a\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=a\\x^2=-2a\left(l\right)\end{cases}}\)
Với \(x^2=a\Rightarrow x^2=\sqrt{x^2-x+1}\)
Đến đây bình phương và tìm ra nghiệm.
KHÓ thật đấy có quản lí mới giải được thôi
Giải các phương trình :
a) \(5+x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=4\left(\sqrt{4-x}+\sqrt{2x-2}\right)\)
b) \(\sqrt[3]{24+x}+\sqrt{12-x}=6\)
\(a,Đk:1\le x\le4\)
Đặt \(y=\sqrt{4-x}+\sqrt{2x-2}\)Ta có: \(y^2=4-x+2x-2+2\sqrt{\left(4-x\right)\left(2x-2\right)}\)
\(\Leftrightarrow x+2+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2\Leftrightarrow x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2-2\)
Phương trình trở thành: \(5+y^2-2=4y\)
\(\Leftrightarrow y^2-4y+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=3\end{cases}}\) ( Vì \(a+b+c=0\))
\(y=1.\) Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=1\Leftrightarrow\sqrt{2x-2}=1-\sqrt{4-x}\)\(\Leftrightarrow\hept{\begin{cases}1-\sqrt{4-x}\ge0\\2x-2=\left(1-\sqrt{4-x}\right)^2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le1\\2x-2=1-2\sqrt{4-x}+4-x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le4-x\le1\\2\sqrt{4-x}=7-3x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}3\le x\le4;7-3x\ge0\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\varnothing\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\) \(\Leftrightarrow x\in\varnothing\)
\(y=3\)Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=3\Leftrightarrow\sqrt{2x-2}=3-\sqrt{4-x}\)\(\Leftrightarrow\hept{\begin{cases}3-\sqrt{4-x}\ge0\\2x-2=\left(3-\sqrt{4-x}\right)^2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2x-2=9-6\sqrt{4-x}+4-x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2\sqrt{4-x}=5-x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}0\le4-x\le9;5-x\ge0\\4\left(4-x\right)=\left(5-x\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\x^2-6x+9=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất là \(x=3\)
(Làm xong hoa mắt :((
Giải phương trình
a)\(\sqrt{2\left(3x-2\right)^2}-5=0\)
b)\(\frac{3}{4}\left(5\sqrt{x+5}-2\right)=\frac{1}{7}\left(6\sqrt{x}+3-10\right)\)
Giải phương trình :
a) \(\left(x-2\right)\left(5-x\right)=7x-\left(x-1\right)\left(3-2x\right)\)
b) \(\sqrt{12-x}+\sqrt[3]{24+x}=6\)
a)\(\left(x-2\right)\left(5-x\right)=7x-\left(x-1\right)\left(3-2x\right)\Leftrightarrow5x-x^2-10+2x=7x-3x+2x^2+3-2x\Leftrightarrow-3x^2+5x-13=0\)\(\Delta=b^2-4ac=25-4.\left(-3\right).\left(-13\right)=-131< 0\)
\(\Rightarrow\)phương trình vô nghiệm
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!
Giải phương trình :
\(\sqrt{x\left(3x+1\right)}-\sqrt{x\left(x-1\right)}=2\sqrt{x^2}\)
dk \(\hept{\begin{cases}x\left(3x+1\right)\ge0\\x\left(x-1\right)\ge0\end{cases}< =>\orbr{\begin{cases}x\ge1\\x\le\frac{-1}{3}\end{cases}}}\)
vì x khác 0 nên chia cả 2 vế cho \(\sqrt{x}\)ta được \(\sqrt{3x+1}-\sqrt{x-1}=2\sqrt{x}< =>\)\(\sqrt{x-1}+2\sqrt{x}-\sqrt{3x+1}=0< =>\)\(\sqrt{x-1}+\frac{4x-\left(3x+1\right)}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(\sqrt{x-1}+\frac{x-1}{2\sqrt{x}+\sqrt{3x+1}}=0\)\(< =>\sqrt{x-1}\left(1+\frac{\sqrt{x-1}}{2\sqrt{x}+\sqrt{3x+1}}\right)=0< =>\sqrt{x-1}=0\) (vì biểu thức trong ngoặc luôn \(\ge1\)) <=> x-1= 0 <=> x=1 (thỏa mãn điều kiện)
Giải phương trình: \(\hept{\begin{cases}\frac{x^3+x^2+x}{x+1}=\left(y+3\right)\sqrt{\left(x+1\right)\left(y+2\right)}\\3x^2-8x-3=4\left(x+1\right)\sqrt{y+2}\end{cases}}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)