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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
Câu 1 :
Xét điều kiện:\(\hept{\begin{cases}x\ge5\\x\le1\end{cases}}\)(Vô lý)
Vậy pt vô nghiệm
Câu 2 :
\(2\sqrt{x+2}+2\sqrt{x+2}-3\sqrt{x+2}=1\)\(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\)
Vậy x=-1
Câu 3 :
\(\sqrt{3x^2-4x+3}=1-2x\)\(\Leftrightarrow3x^2-4x+3=1+4x^2-4x\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt{2}\)
Câu 4 :
\(4\sqrt{x+1}-3\sqrt{x+1}=4\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x=15\)
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
câu 2 ĐK \(x\ge1\)
\(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+18-\left(x+26\right)\sqrt{x-1}=0\)
<=> \(\left(5x+8\right)\left(\sqrt{2x-1}-1\right)+7x\left(\sqrt{x+3}-2\right)+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)
<=>\(\left(5x+8\right).\frac{2x-2}{\sqrt{2x-1}+1}+7x.\frac{x+3-4}{\sqrt{x+3}+2}+\left(x+26\right)\sqrt{x-1}+10\left(x-1\right)=0\)
<=> \(\sqrt{x-1}\left(\frac{2\left(5x+8\right)\sqrt{x-1}}{\sqrt{2x-1}+1}+\frac{7x\sqrt{x-1}}{\sqrt{x+3}+2}+\left(x+26\right)+10\sqrt{x-1}\right)=0\)
Với \(x\ge1\)thì cái trong ngoặc >0
=> \(x=1\)
Vậy x=1
Câu 3 . Bạn chuyển vế rồi lập phương là ra ngay.
3) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
<=> \(\sqrt[3]{3x-1}+\sqrt[3]{5-x}=-\sqrt[3]{2x-9}+\sqrt[3]{4x-3}\)
<=> \(3x+1+5-x+3\sqrt[3]{\left(3x+1\right)^2}.\sqrt[3]{5-x}+3\sqrt[3]{\left(3x+1\right).\left(5-x\right)}\)\(=4x-3-2x+9-3\sqrt[3]{\left(4x-3\right)^2.\left(2x-9\right)}+3\sqrt[3]{\left(4x-3\right).\left(2x-9\right)^2}\)
<=> \(\sqrt[3]{\left(3x+1\right).\left(5-x\right)}.\left(\sqrt[3]{3x+1}+\sqrt[3]{5-x}\right)\) \(=\sqrt[3]{\left(4x-3\right).\left(2x-9\right)}.\left(\sqrt[3]{2x-9}-\sqrt[3]{4x-3}\right)\)
<=> \(\sqrt[3]{\left(3x+1\right).\left(5-x\right)}=-\sqrt[3]{\left(4x-3\right).\left(2x-9\right)}\)
<=> (3x+`1).(5-x ) = - ( 4.x - 3 ) . ( 2x - 9 )
<=> -3.x2 -x + 15x + 5 = -8.x2 + 36x + 6x -27
<=> 5x2 -28x + 32 = 0
<=> \(\orbr{\begin{cases}x=4\\x=\frac{8}{5}\end{cases}}\)
Vậy x = \(\left\{4;\frac{8}{5}\right\}\)
4) Đặt : a = \(\sqrt{4-x}\left(a\ge0\right)\Leftrightarrow x=4-a^2\)
: b = \(\sqrt{x+1}\left(b\ge0\right)\Leftrightarrow b^2=x+1=5-a^2\)
<=> a2 + b2 = 5
Phương trình trở thành : ( b2 + 16 ) .a + ( 16 + a2 ) .b -9ab = 36
<=> a2b + ab2 + 16. ( a + b ) - 9ab = 36
<=> ( a + b ) .( ab +16 ) = 36 + 9ab = 9 .( 4 + ab )
<=> ( a2 + b2 + 2ab ) . ( a2 b2 + 32ab + 256 ) = 81 . ( 16 + 8ab +a2 b2 )
\---------/
5
Thay vào ta được : 5a2 b2 + 672ab + 2a3 b3 + 64a2b2 + 1280=1296 + 648ab + 81a2b2
<=> 2a3b3 - 12.a2b2 + 24ab - 16 = 0 <=> ab = 2 (1)
Từ ( 1 ) => ta có được : \(\sqrt{4-x}.\sqrt{x+1}=2\)
<=> ( 4 - x ). ( x + 1 ) = 4
<=> -x2 + 3x + 4 = 4
<=> -x2 + 3x = 0
<=>\(\orbr{\begin{cases}x=0\\x=3\end{cases}\left(T/m\right)}\)
Vậy nghiệm cuối cùng là :{ 0 ; 3 }
Trl :
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Trl:
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