Gpt: \(\sqrt{x^2-4x+4}=3\)
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gpt:
\(x^2+2x+4=3\sqrt{x^3+4x}\)
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow x^2+4+2x=3\sqrt{x\left(x^2+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{x^2+4}=b>0\end{matrix}\right.\)
\(\Rightarrow b^2+2a^2=3ab\)
\(\Leftrightarrow2a^2-3ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=2a\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{x^2+4}\\\sqrt{x^2+4}=2\sqrt{x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2-4x+4=0\end{matrix}\right.\) \(\Rightarrow x=2\)
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GPT:
1/ \(\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1\)
2/ \(\sqrt[3]{\frac{12x^2+12x+9}{4}}=x+\sqrt[4]{\frac{4x^3-2}{3}}\)
GPT:\(\sqrt{1-x}+\sqrt{4+x}=3...\)
\(x^2+4x+5=2\sqrt{2x+3}\)
GPT :\(\sqrt{x^2-3x+2}\) +\(\sqrt{x^2-4x+3}\) =\(2\sqrt{x^2-5x+4}\)
mk cx toán nek
câu này cx bình thường, bn cố nhìn ik , ra ngay thôi, mk mún bn tự suy nghĩ tư duy
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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
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Gpt :
1) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
2) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+s}+\sqrt{x+1}=16\)
3)\(\sqrt{4x+20}+\sqrt{x+5}-\frac{1}{3}\sqrt{9x+45}=4\)
4) \(\frac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\)
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
\(4\sqrt{x+4}+\sqrt{16-3x}=x^2+4x+12\)
GPT: \(\sqrt{x^2+4x+12}=2x-4+\sqrt{x+1}\)
ĐLXĐ:\(x\ge-1\)
\(\sqrt{x^2+4x+12}=2x-4+\sqrt{x+1}\)
\(\Leftrightarrow\left[\sqrt{x^2+4x+12}-\left(6-3x\right)\right]-\left[\sqrt{x+1}-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\frac{x^2+4x+12-36+36x-9x^2}{\sqrt{x^2+4x+12}+2-3x}-\frac{x+1-x^2+4x-4}{\sqrt{x+1}+x+2}=0\)
\(\Leftrightarrow\frac{-8x^2+40x-24}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)
\(\Leftrightarrow\frac{8\left(-x^2+5x-3\right)}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)
\(\Leftrightarrow\left(-x^2+5x-3\right)\left[\frac{8}{\sqrt{x^2+4x+12}+2-3x}-\frac{1}{\sqrt{x+1}+x-2}\right]=0\)
TH1:\(-x^2+5x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)
TH2:........ ( chắc vô nghiệm )
phần mẫu phải là \(\sqrt{x^2+4x+12}+6-3x\) chứ :vv Hơi lỗi nhưng cảm ơn nhé !!
\(x^2+4x+12=\left(x+1\right)^2+2\left(x+1\right)+9\)
Đặt \(\sqrt{x+1}=a\ge0\).
PT \(\Leftrightarrow\sqrt{a^4+2a^2+9}=2a^2+a-6\)