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a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
h)
ĐK: \(\left\{\begin{matrix} 3x-12\geq 0\\ x-5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 4\\ x\neq 5\end{matrix}\right.\)
k)
ĐK: \(\left\{\begin{matrix} x-1\geq 0\\ x-2\neq 0\\ x-3\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\neq 2\\ x\neq 3\end{matrix}\right.\)
m)
ĐK: \(\left\{\begin{matrix} x-2\neq 0\\ x-4\neq 0\\ \frac{2x-3}{x-2}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 2\\ x\neq 4\\ x>2\end{matrix}\right.\) hoặc \(x\leq \frac{3}{2}\)
Lời giải:
a) ĐK: $-4x+16\geq 0\Leftrightarrow x\leq 4$
b) ĐK: \(\left\{\begin{matrix} 2x-1\neq 0\\ \frac{-3}{2x-1}\geq 0\end{matrix}\right.\Leftrightarrow 2x-1< 0\Leftrightarrow x< \frac{1}{2}\)
c) ĐK: $-5x^2\geq 0\Leftrightarrow 5x^2\leq 0$. Mà $5x^2\geq 0$ với mọi $x\in\mathbb{R}$ nên biểu thức có nghĩa khi $5x^2=0\Leftrightarrow x=0$
d) ĐK:
\(\left\{\begin{matrix} -x^2-4x-4\neq 0\\ \frac{-3}{-x^2-4x-4}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -(x+2)^2\neq 0\\ \frac{3}{(x+2)^2}\geq 0\end{matrix}\right.\Leftrightarrow x\neq -2\)
e) ĐK: $\frac{2x-4}{-3}\geq 0\Leftrightarrow 2x-4\leq 0\Leftrightarrow x\leq 2$
f) ĐK: \(\left\{\begin{matrix} 3x-9\geq 0\\ 2x-8>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ x>4\end{matrix}\right.\Leftrightarrow x>4\)
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
a) ĐKXĐ: \(5x-7\ge0\) \(\Leftrightarrow\)\(x\ge\frac{7}{5}\)
b) ĐKXĐ: \(2x^2+x\ge0\)\(\Leftrightarrow\) \(x\left(2x+1\right)\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x\ge0\\x\le-\frac{1}{2}\end{cases}}\)
c) ĐKXĐ: \(4-7x\ge0\)\(\Leftrightarrow\)\(x\le\frac{4}{7}\)
d) ĐKXĐ: \(x^3+x\ge0\) \(\Leftrightarrow\)\(x\left(x^2+1\right)\ge0\)\(\Leftrightarrow\)\(x\ge0\)
e) ĐKXĐ: \(\frac{x-5}{2x+1}\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x\ge5\\x< -\frac{1}{2}\end{cases}}\)
f) ĐKXĐ: \(\frac{3-2x}{3x-2}\ge0\) \(\Leftrightarrow\)\(\frac{2}{3}< x\le\frac{3}{2}\)