![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có \(\left(x+3\right)\sqrt{15-x^2}=\left(x-3\right)\left(x+4\right)\)
<=> \(\left(x-3\right)\left(\sqrt{15-x^2}-x-4\right)=0\)
đến đây dễ rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
Đkxđ: \(x\ge3\)
pt đã cho \(\Leftrightarrow x^2-x-12+3\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)+3.\dfrac{x-4}{\sqrt{x-3}+1}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3+\dfrac{3}{\sqrt{x-3}+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x+3+\dfrac{3}{\sqrt{x-3}+1}=0\left(vôlí\right)\end{matrix}\right.\)
Vậy pt đã cho có nghiệm duy nhất \(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x-3}+\sqrt{4x-12}=6\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}+2\sqrt{x-3}=6\)
\(\Leftrightarrow3\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=2\)
\(\Leftrightarrow x-3=4\Leftrightarrow x=7\left(tm\right)\)
\(\sqrt{x-3}+\sqrt{4x-12}=6\)đk : x >= 3
\(\Leftrightarrow\sqrt{x-3}+2\sqrt{x-3}=6\Leftrightarrow3\sqrt{x-3}=6\Leftrightarrow\sqrt{x-3}=2\)
\(\Leftrightarrow x-3=4\Leftrightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt $\sqrt[3]{x}=a; \sqrt[3]{2x-3}=b$. Ta có:
\(\left\{\begin{matrix} a+b=\sqrt[3]{4(a^3+b^3)}\\ 2a^3-b^3=3\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} a^3+b^3+3ab(a+b)=4(a^3+b^3)\\ 2a^3-b^3=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^3+b^3=ab(a+b)\\ 2a^3-b^3=3\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} (a-b)^2(a+b)=0(1)\\ 2a^3-b^3=3(2)\end{matrix}\right.\)
Từ $(1)$ suy ra $a=b$ hoặc $a=-b$.
Nếu $a=b$. Thay vào $(2)$ suy ra $a^3=b^3=3$
$\Leftrightarrow x=2x-3=3$ (thỏa mãn)
Nếu $a=-b$. Thay vào $(2)$ suy ra $a^3=1; b^3=-1$
$\Leftrightarrow x=1; 2x-3=-1$ (thỏa mãn)
Vậy $x=3$ hoặc $x=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt[3]{x}+\sqrt[3]{2x-3}=\sqrt[3]{12\left(x-1\right)}\left(1\right)\)
\(\left(1\right)\Leftrightarrow x+2x-3+3.\sqrt[3]{x\left(2x-3\right)}.\left(\sqrt[3]{x}+\sqrt[3]{2x-3}\right)=12x-12\)
\(\Rightarrow\sqrt[3]{12x\left(x-1\right)\left(2x-3\right)}=3x-3\)
\(\Leftrightarrow12x\left(x-1\right)\left(2x-3\right)=[3\left(x-1\right)]^3\)
\(\Leftrightarrow12x\left(2x^2-5x+3\right)=27\left(x^3-3x^2+3x-1\right)\)
\(\Leftrightarrow24x^3-60x^2+36x=27x^3-81x^2+81x-27\)
\(\Leftrightarrow3x^3-21x^2+45x-27=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\). Thử lại ta thấy cả x=1;x=3 đều t/m bài toán
Vậy, pt có tập nghiệm là S={1;3}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}=\sqrt{3}-x^2\)
\(\Leftrightarrow\sqrt{-x^2+4x+12}=\sqrt{3}-x^2+\sqrt{-x^2+2x+3}\)
\(VP=\sqrt{-x^2+4x+12}=\sqrt{-\left(x-2\right)^2+16}\le4\)
\(VT=\sqrt{3}-x^2+\sqrt{-x^2+2x+3}=\sqrt{3}-x^2+\sqrt{-\left(x-1\right)^2+4}\)
\(\le\sqrt{3}+2
Dat \(\sqrt[3]{12-x}=a;\)\(\sqrt[3]{x+15}=b\)
Khi do ta co: \(\hept{\begin{cases}a+b=3\\a^3+b^3=27\end{cases}}\) <=> \(\hept{\begin{cases}a=3-b\\a^3+b^3=27\end{cases}}\) <=> \(\hept{\begin{cases}a=3-b\\\left(3-b\right)^3+b^3=27\end{cases}}\)
<=> \(\hept{\begin{cases}a=3-b\\9\left(b^2-3b+3\right)=27\end{cases}}\) <=> \(\hept{\begin{cases}a=3-b\\b^2-3b+3=3\end{cases}}\) <=> \(\hept{\begin{cases}a=3-b\\b\left(b-3\right)=0\end{cases}}\)
Xet: \(b\left(b-3\right)=0\)
<=> \(\orbr{\begin{cases}b=0\\b=3\end{cases}}\)
Đến đây tự giải