Giai phuong trinh : \(\left(x+5\right)\sqrt{\left(x+1\right)+1}=\sqrt[3]{\left(3x+4\right)}\)
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Giai phuong trinh giup minh 3 cau nay voi
a,\(3x\left(2-\sqrt{4}\right)=3\left(\sqrt{4}x+1\right)\)
b,\(\left(5-x\right).\left(\sqrt{3}+x\right)-5=0.\)
c,\(\left(x^2-2x\right)+\left(-4+8x\right)=0.\)
giai phuong trinh
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
ĐK: \(x\ge-2\)
\(pt\Leftrightarrow\frac{x+5-\left(x+2\right)}{\sqrt{x+5}+\sqrt{x+2}}.\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)=3\)
\(\Leftrightarrow3.\frac{1+\sqrt{x+2}.\sqrt{x+5}}{\sqrt{x+2}+\sqrt{x+5}}=3\)
\(\Leftrightarrow1+\sqrt{x+2}\sqrt{x+5}=\sqrt{x+2}+\sqrt{x+5}\)
\(\Leftrightarrow\left(\sqrt{x+2}-1\right)\left(\sqrt{x+5}-1\right)=0\)
\(\Leftrightarrow\sqrt{x+2}=1\text{ hoặc }\sqrt{x+5}=1\)
\(\Leftrightarrow x=-1\text{ (nhận) hoặc }x=-4\text{ (loại)}\)
Vậy tập nghiệm của pt là: \(S=\left\{1\right\}\)
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Giai phuong trinh va he phuong trinh:
a) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
b) \(x^2+3x+1=\left(x+3\right).\sqrt{x^2+1}\)
c) \(\left\{{}\begin{matrix}x^2+y^2=11\\x+xy+y=3+4\sqrt{2}\end{matrix}\right.\)
tim nghiem nguyen cua phuong trinh: \(\left(x^2+1\right)\sqrt{1-x}-\left(2x+x^3\right)\sqrt{x+1}=3x^4\sqrt{2x}\)
Giai phuong trinh
\(\sqrt{x.\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2x\)
x=0
nhanh nhất có thể
X=0
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giai phuong trinh
\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(\Leftrightarrow\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
\(\Rightarrow\left(x-2\right)^2=x^2-4\)
\(\Leftrightarrow x^2-4x+4-x^2+4=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow x=2\)
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Đặt \(\sqrt[3]{x+2}=a;\sqrt[3]{x-2}=b;\) ta có:
\(2a^2-b^2=ab\) ⇔ \(2a^2-ab-b^2=0\)
\(\Leftrightarrow2a^2+ab-2ab-b^2=0\)
⇔ \(\left(2a+b\right)\left(a-b\right)=0\)
⇔ \(\left[{}\begin{matrix}2\sqrt[3]{x+2}=-\sqrt[3]{x-2}\\\sqrt[3]{x-2}=\sqrt[3]{x+2}\end{matrix}\right.\)⇔ \(x=-\frac{14}{9}\)
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Giai phuong trinh:
\(28+\sqrt[3]{x^2}=3x+2\sqrt[3]{x}+\left(x-4\right)\sqrt{x-7}\)
Giai he phuong trinh
1) \(\hept{\begin{cases}\left(x^4+1\right)\left(y^4+1\right)=4xy\\\sqrt[3]{x-1}-\sqrt{y-1}=1-x^3\end{cases}}\)
2) \(\hept{\begin{cases}\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)=2012\\x^2+z^2-4\left(y+z\right)+8=0\end{cases}}\)
2)
sử dụng phương pháp nhân liên hợp ở pt (1) ta được
\(\hept{\begin{cases}x+\sqrt{2012+x^2}=\sqrt{y^2+2012}-y\\y+\sqrt{y^2+2012}=\sqrt{x^2+2012}-x\end{cases}}\)
cộng 2 vế lại được x=-y
rồi sao?? mik đíu hiểu pt 2 lôi z ở đâu
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2,RA DUOC X=-Y ...THAY VAO PT 2 TA DC Y^2+Z^2 -4Y-4Z +4+4=0...(Y-2)^2 +(Z-2)^2=0...Y=Z=2 , X=-Y=-2
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Giai phuong trinh
1/ sqrt{x^2+4x+5}+sqrt{x^2-6x+13}3
2/ sqrt{3x^2-18x+28}+sqrt{4x^2-24x+45}6x-x^2-5
3/ sqrt{2x^2-4x+27}+sqrt{3x^2-6x+12}4x^2+8x+4
4/ sqrt{x^2+x+7}+sqrt{x^2+x+2}sqrt{3x^2+3x+19}
5/ left(x+2right)left(x+3right)-sqrt{x^2+5x+1}9
6/ left(x+4right)left(x+1right)-3sqrt{x^2+5x+2}6
7/ sqrt{2x^2+3x+5}+sqrt{2x^2-3x+5}3sqrt{x}
Đọc tiếp
Giai phuong trinh
1/ \(\sqrt{x^2+4x+5}+\sqrt{x^2-6x+13}=3\)
2/ \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=6x-x^2-5\)
3/ \(\sqrt{2x^2-4x+27}+\sqrt{3x^2-6x+12}=4x^2+8x+4\)
4/ \(\sqrt{x^2+x+7}+\sqrt{x^2+x+2}=\sqrt{3x^2+3x+19}\)
5/ \(\left(x+2\right)\left(x+3\right)-\sqrt{x^2+5x+1}=9\)
6/ \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
7/ \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3\sqrt{x}\)
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 ....
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