\(Gpt:\sqrt{x^4-7}+\sqrt{x^3-7}=x^2\)
Những câu hỏi liên quan
GPt \(\sqrt{x+2}-\sqrt{x+3}=\sqrt{x+4}-\sqrt{x+7}\)
gpt \(\sqrt{x+8+2\sqrt{x+7}}+\sqrt{x+1-\sqrt{x+7}}=4\)
\(\sqrt{x+2\sqrt{x-2}}-\sqrt{x-2\sqrt{x-2}}=-2\)
\(=\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)ĐK:\(x\ge-7\)
Đặt \(t=\sqrt{x+7}\left(t\ge0\right)\)
\(\Rightarrow t+1-4=\sqrt{t^2-t-6}\)
\(\Leftrightarrow t^2-6t+9=t^2-t-6\left(t\ge3\right)\)
\(\Leftrightarrow5t=15\)
\(\Leftrightarrow t=3\left(TM\right)\)\(\Rightarrow x=2\left(tm\right)\)
S={2}
b)ĐK:\(x\ge2\)
pt\(\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+2}-\sqrt{x-2-2\sqrt{x-2}+2}=-2\)
Đặt t= can(x-2)(t>=0)
Đến đây bạn giải tiếp nhé!
#Walker
12 tháng 12 2021 lúc 9:56
Gpt: \(5\sqrt{x-1}-\sqrt{x+7}=3x-4\) (2 cách)
Cách 1:
GPT :\(5\sqrt{x-1}-\sqrt{x+7}=3x-4\) - Hoc24
Cách 2:
Đặt \(\left\{{}\begin{matrix}\sqrt{25x-25}=a\\\sqrt{x+7}=b\end{matrix}\right.\) \(\Rightarrow3x-4=\dfrac{a^2-b^2}{8}\)
Pt trở thành:
\(a-b=\dfrac{a^2-b^2}{8}\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-8\right)=0\)
\(\Leftrightarrow...\)
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gpt \(\sqrt{\frac{x}{3}-3}+\sqrt{7-\frac{x}{3}}=2x-7-\frac{x^2}{9}\)
Gpt:
\(4\sqrt{x+3}+\sqrt{x-1}=x+7\)
14 tháng 1 2021 lúc 12:59
Giải phương trình 1, \(x^2+9x+7=\left(2x+1\right)\sqrt{2x^2+4x+5}\)
2, GPT \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
3. GHPT \(\left\{{}\begin{matrix}x^2-2y-1=2\sqrt{5y+8}+\sqrt{7x-1}\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
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2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
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3.
ĐKXĐ: ...
Từ pt dưới:
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow y=x-2\)
Thế vào pt trên:
\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)
\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)
\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)
\(\Leftrightarrow x^2-5x+2=0\)
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GPT:
\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}-3\sqrt{3}x=6\sqrt{3}\)
cai nay la hag dag thuc phan tih ra la dk
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pt<=>căn((x-1/2)^2+75/4)+căn(2(x-1/2)^2+3(x+2)^2)+căn((x-1/2)^2+3(2x+3/2)^2)>=3*căn3(x+2)
dấu = xãy ra khi x=1/2
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1.Gpt: \(\dfrac{6}{x-3\sqrt{x-2}+7}=\dfrac{1}{\sqrt{x-2}}+\dfrac{\sqrt{3}}{3\sqrt{2\sqrt{x-2}}-3}\)
2.Ghpt: \(\left\{{}\begin{matrix}x^2-y-z=0\\x^3-y^2-z^2+2=0\end{matrix}\right.\)
gpt: \(2\sqrt{3x+7}-5\sqrt[3]{x-6}=4\)
\(\left(x^2-3x+2\right)\left(x^2-12x+32\right)\le4x^2\)
\(\left(\sqrt{x+1}-1\right)\left(\sqrt{x^2-4x+7}+1\right)=x\)