Giải phương trình:
a, \(\sqrt[3]{x^2-1}+x=\sqrt{x^3-2}\)
b, \(x\sqrt[3]{35-x^3}.\left(x+\sqrt[3]{35-x^3}\right)=30\)
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a)iải phương trình sau: - K2PI – TOÁN THPT | Chia sẻ Tài liệu, đề thi, hỗ trợ giải toán
b)giải pt: x^2 + 3x+1=(x+3)căn(x^2+1)? | Yahoo Hỏi & Đáp
c)chuyển vế bình
a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)
Vậy ...
\(\left(x+\sqrt[3]{35-x^3}\right)^3=x^3+35-x^3+3x\sqrt[3]{35-x^3}\left(x+\sqrt[3]{35-x^3}\right)=35+3.30=125\\ \Leftrightarrow x+\sqrt[3]{35-x^3}=5\\ \Leftrightarrow x-5=-\sqrt[3]{35-x^3}\)Bạn lập phương 2 vế rồi giải bình thường nhé.
b.
ĐKXĐ: \(x\ge-1\)
\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)
\(\Leftrightarrow...\)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+2a^2=-b^2+b+3ab\)
\(\Leftrightarrow\left(2a^2-3ab+b^2\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a+1=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x+5+4\sqrt{x+1}=1-x\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)
\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)
\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)
a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)
\(\Leftrightarrow7\left|x-1\right|=35\)
\(\Leftrightarrow\left|x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)
c) ĐKXĐ: \(x\ge0\)
Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
\(\Leftrightarrow x-1=x+\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}-6=-1\)
\(\Leftrightarrow\sqrt{x}=5\)
hay x=25(nhận)
a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
a/ ĐKXĐ: \(x\ge\sqrt[3]{2}\)
\(\Leftrightarrow\sqrt{x^3-2}-\left(2x-1\right)+x-1-\sqrt[3]{x^2-1}=0\)
\(\Leftrightarrow\frac{x^3-2-\left(2x-1\right)^2}{\sqrt{x^3-2}+2x-1}+\frac{\left(x-1\right)^3-\left(x^2-1\right)}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}=0\)
\(\Leftrightarrow\frac{x^3-4x^2+4x-3}{\sqrt{x^3-2}+2x-1}+\frac{x^3-4x^2+3x}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x^2-x+1\right)}{\sqrt{x^3-2}+2x-1}+\frac{\left(x-3\right)\left(x^2-x\right)}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{x^2-x+1}{\sqrt{x^3-2}+2x-1}+\frac{x^2-x}{\left(x-1\right)^2+\left(x-1\right)\sqrt[3]{x^2-1}+\sqrt[3]{\left(x^2-1\right)^2}}\right)=0\)
\(\Rightarrow x=3\)
b/ Đặt \(\sqrt[3]{35-x^3}=a\)
\(\Rightarrow\left\{{}\begin{matrix}ax\left(a+x\right)=30\\x^3+a^3=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ax\left(a+x\right)=90\\x^3+a^3=35\end{matrix}\right.\)
\(\Rightarrow x^3+a^3+3ax\left(a+x\right)=125\)
\(\Leftrightarrow\left(x+a\right)^3=125\)
\(\Leftrightarrow x+a=5\)
\(\Leftrightarrow a=5-x\)
\(\Leftrightarrow\sqrt[3]{35-x^3}=5-x\)
\(\Leftrightarrow35-x^3=125-75x+15x^2-x^3\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow...\)