\(\sqrt[3]{35-x^3}\left(x+\sqrt[3]{35-x^3}\right)=30\)
Những câu hỏi liên quan
\(x\sqrt[3]{35-x^3}\times\left(x+\sqrt[3]{35-x^3}\right)=30\)
\(CMR:A=\frac{1}{\left(\sqrt{1}+\sqrt{3}\right)^3}+\frac{1}{\left(\sqrt{3}+\sqrt{5}\right)^3}+...+\frac{1}{\left(\sqrt{2003}+\sqrt{2005}\right)^3}< \frac{246}{2007}\)
Giải phương trình: \(x\sqrt[3]{35-x^3}\left(x+\sqrt[3]{35-x^3}\right)=30\)
\(\left(x+\sqrt[3]{35-x^3}\right)^3=x^3+35-x^3+3x\sqrt[3]{35-x^3}\left(x+\sqrt[3]{35-x^3}\right)=35+3.30=125\\ \Leftrightarrow x+\sqrt[3]{35-x^3}=5\\ \Leftrightarrow x-5=-\sqrt[3]{35-x^3}\)Bạn lập phương 2 vế rồi giải bình thường nhé.
Đúng 0
Bình luận (0)
giải phương trình
\(x\sqrt[3]{35-x^3}\left(x+\sqrt[3]{35-x^3}\right)=30\)
\(^{x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}}\)
\(x+\sqrt{x+\sqrt{x-1}}=6\)
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
Đúng 0
Bình luận (0)
Giải phương trình: \(x\sqrt[3]{35-x^2}\left(x+\sqrt[3]{35-x^2}\right)=30\)
Giải phương trình:
a, \(\sqrt[3]{x^2-1}+x=\sqrt{x^3-2}\)
b, \(x\sqrt[3]{35-x^3}.\left(x+\sqrt[3]{35-x^3}\right)=30\)
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 \(\sqrt[3]{35-x^3}=a\)
\(\Rightarrow\left\{{}\begin{matrix}ax\left(a+x\right)=30\\x^3+a^3=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3ax\left(a+x\right)=90\\x^3+a^3=35\end{matrix}\right.\)
\(\Rightarrow x^3+a^3+3ax\left(a+x\right)=125\)
\(\Leftrightarrow\left(x+a\right)^3=125\)
\(\Leftrightarrow x+a=5\)
\(\Leftrightarrow a=5-x\)
\(\Leftrightarrow\sqrt[3]{35-x^3}=5-x\)
\(\Leftrightarrow35-x^3=125-75x+15x^2-x^3\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow...\)
giaỉ pt:
a, \(\sqrt{x +1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\)
b, \(14\sqrt{x+35}+6\sqrt{x+1}=84+\sqrt{x^2+36x+35}\)
c, \(x\sqrt{2x+3}+3\left(\sqrt{x+5}+1\right)=3x+\sqrt{2x^2+13x+15}+\sqrt{2x+3}\)
b.
ĐKXĐ: \(x\ge-1\)
\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)
\(\Leftrightarrow...\)
Đúng 2
Bình luận (0)
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) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)
\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)
\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
c.
ĐKXĐ: \(x\ge-\dfrac{3}{2}\)
\(\Leftrightarrow x\sqrt{2x+3}-\sqrt{2x+3}+3-3x+3\sqrt{x+5}-\sqrt{\left(2x+3\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\sqrt{2x+3}\left(x-1\right)-3\left(x-1\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\sqrt{2x+3}-3\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left(x-1-\sqrt{x+5}\right)\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{x+5}=0\\\sqrt{2x+3}-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5-\sqrt{x+5}-6=0\\\sqrt{2x+3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-2\left(loại\right)\\\sqrt{x+5}=3\\\sqrt{2x+3}=3\end{matrix}\right.\)
\(\Leftrightarrow...\)
Đúng 1
Bình luận (0)
\(X\cdot\sqrt[3]{35-X^3}\cdot\left(X+\sqrt[3]{35-X^3}\right)=5\)
\(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)
b) \(x^2+8x-3=2\sqrt{x\left(8+x\right)}\)
c)\(\sqrt{x-2}+\sqrt{x+1}+\sqrt{2x+3}=6\)
Tìm x
1, 4x^4+12x^3+12x-47x^2+404x^4+12x^3+12x-47x^2+40
2, x^2+sqrt{x+1}1
3.sqrt{2x^2+8x+6}+sqrt{x^2-1}2x+2
4.x-2sqrt{x-1}-left(x-1right)sqrt{x}+sqrt{x^2-x}0
5.xsqrt[3]{35-x^3}-left(x+sqrt[3]{35-x^3}right)30
6. x^3-12sqrt[3]{2x-1}
Đọc tiếp
Tìm x
1, \(4x^4+12x^3+12x-47x^2+4=0\)\(4x^4+12x^3+12x-47x^2+4=0\)
2, \(x^2+\sqrt{x+1}=1\)
3.\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
4.\(x-2\sqrt{x-1}-\left(x-1\right)\sqrt{x}+\sqrt{x^2-x}=0\)
5.\(x\sqrt[3]{35-x^3}-\left(x+\sqrt[3]{35-x^3}\right)=30\)
6. \(x^3-1=2\sqrt[3]{2x-1}\)
1/\(4x^4+12x^3-47x^2+12x+4=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^3+20x^2-7x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(2x^2+11x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=\frac{-11\pm\sqrt{105}}{4}\end{matrix}\right.\)
Vậy ....
1, 4x^4+12x^3+12x−47x^2+4=0 nhé