Giải pt
\(x\sqrt[3]{9-x^3}\left(x+\sqrt[3]{9-x^3}\right)=6\)
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Những câu hỏi liên quan
27 tháng 11 2021
\(a,ĐK:1\le x\le3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{3-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow a+b-ab=1\Leftrightarrow a+b-ab-1=0\\ \Leftrightarrow\left(a-1\right)\left(1-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=1\\3-x=1\end{matrix}\right.\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:0\le x\le9\\ PT\Leftrightarrow9+2\sqrt{x\left(9-x\right)}=-x^2+9x+9\\ \Leftrightarrow2\sqrt{-x^2+9x}-\left(-x^2+9x\right)=0\\ \Leftrightarrow\sqrt{-x^2+9x}\left(2-\sqrt{-x^2+9x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x^2+9x=0\\\sqrt{-x^2+9x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\\x^2-9x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=9\left(n\right)\\x=\dfrac{9+\sqrt{65}}{2}\left(n\right)\\x=\dfrac{9-\sqrt{65}}{2}\left(n\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
b.
Đặt \(\sqrt[3]{3x-2}=y\Rightarrow y^3=3x-2\)
Ta được hệ:
\(\left\{{}\begin{matrix}x^3+2=3y\\y^3=3x-2\end{matrix}\right.\)
Trừ vế cho vế:
\(x^3-y^3+2=3y-3x+2\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+3\right)=0\)
\(\Leftrightarrow x=y\)
\(\Leftrightarrow x^3=3x-2\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)=0\)
Kiểm tra lại đề câu c
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2021
a.
Đặt \(\sqrt[3]{4x-3}=y\Rightarrow y^3=4x-3\)
Ta được hệ:
\(\left\{{}\begin{matrix}x^3+3=4y\\y^3=4x-3\end{matrix}\right.\)
Trừ về cho vế:
\(x^3-y^3+3=4y-4x+3\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+4\right)=0\)
\(\Leftrightarrow x=y\)
\(\Leftrightarrow x=\sqrt[3]{4x-3}\)
\(\Leftrightarrow x^3-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-3\right)=0\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
L
1
4 tháng 9 2023
\(\sqrt{4x^2}=3\left(ĐK:4x^2\ge0\forall x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x\right)^2}=3\\ \Leftrightarrow\left|2x\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{2};\dfrac{3}{2}\right\}\)
\(\sqrt{x^2-6x+9}=2\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=2\left(ĐK:\left(x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|x-3\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+3\\x=-2-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left(\pm5\right)\)
\(\sqrt{\left(2x-3\right)^2}=6\left(ĐK:\left(2x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|2x-3\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3+6\\2x=-6+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4,5\left(tm\right)\\x=-1,5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{4,5;-1,5\right\}\)
\(\sqrt{25x^2}=100\\ \sqrt{\left(5x\right)^2}=100\left(ĐK:\left(5x\right)^2\ge0\forall x\in R\right)\\\Leftrightarrow \left|5x\right|=100\\ \Leftrightarrow\left[{}\begin{matrix}5x=100\\5x=-100\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=20\left(tm\right)\\x=-20\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\pm20\right\}\)
AH
Akai Haruma
Giáo viên
25 tháng 5 2023
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 4$
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-1\right]:\left[\frac{(3-\sqrt{x})(3+\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+3)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right]\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{-(\sqrt{x}-2)}=\frac{3}{\sqrt{x}-2}\)
LT
1
FM
24 tháng 9 2018
ĐKXĐ: x\(\ge0\)
Ta có:
\(\left(x+3\sqrt{x}\right)\left(x+9\sqrt{x}+18\right)=168\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x+3}\right)\left(\sqrt{x}+6\right)=168\)
\(\Leftrightarrow\left(x+6\sqrt{x}\right)\left(x+6\sqrt{x}+9\right)=168\)
Đặt \(x+6\sqrt{x}=a\)\(\left(a\ge o\right)\). Khi đó:
\(a\left(a+9\right)=168\Leftrightarrow a^2+9a-168=0\)
Bn tu giải tiếp nhé
Đặt \(a=x\\ b=\sqrt[3]{9-x^3}\\ \Rightarrow a^3+b^3=9\)
\(ab\left(a+b\right)=6\\ \Rightarrow a+b=\frac{6}{ab}\)
Mà ta có \(a^3+b^3=9\\ \Rightarrow\left(a+b\right)\left(a^2+b^2-ab\right)=9\\ \Rightarrow a+b=\frac{9}{a^2+b^2-ab}\)
\(\Rightarrow\frac{6}{ab}=\frac{9}{a^2+b^2-ab}\\ \Rightarrow6a^2+6b^2-15ab=0\\ \Rightarrow\left(6a-3b\right)\left(a-2b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=\frac{1}{2}b\\a=2b\end{cases}\Rightarrow\orbr{\begin{cases}2x=\sqrt[3]{9-x^3}\\x=2\sqrt[3]{9-x^3}\end{cases}\Rightarrow}\orbr{\begin{cases}8x^3=9-x^3\\x^3=72-8x^3\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Đặt \(\sqrt[3]{9-x^3}=a.\)