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a) ĐKXĐ: \(\left\{{}\begin{matrix}5-x\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x\ge-5\\x\ge3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ge3\end{matrix}\right.\Leftrightarrow3\le x\le5\)
Ta có: \(\sqrt{5-x}+\sqrt{x-3}=\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x-3}\right)^2=\left(\sqrt{2}\right)^2\)
\(\Leftrightarrow5-x+2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}+x-3=2\)
\(\Leftrightarrow2+2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}=2\)
\(\Leftrightarrow2\cdot\sqrt{\left(5-x\right)\cdot\left(x-3\right)}=0\)
mà \(2\ne0\)
nên \(\sqrt{\left(5-x\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\left(5-x\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
Vậy: S={3;5}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow x-2\ge0\)\(\Leftrightarrow x\ge2\)
Ta có: \(\sqrt{x^2-4}=2\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{x-2}\cdot\sqrt{x+2}-2\cdot\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\cdot\left(\sqrt{x+2}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\)
Vậy: S={2}
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
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
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/ 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:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
\(\Leftrightarrow\sqrt{x-2}-1+\sqrt{4-x}-1=2x^2-5x-3\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}+\frac{1}{\sqrt{4-x}+1}+2x+1\right)=0\)
\(\Rightarrow x=3\)
phương trình còn lại mk chưa giải đc nhưng nó vô nghiệm
Em thử câu c nha, sai thì thôi
c) ĐK: \(x\ge-1\).Nhận xét x = 0 là không phải nghiệm, xét x khác 0:
Nhân liên hợp ta được \(\left(x+4\right).\left(\frac{x}{\sqrt{x+1}-1}\right)^2=x^2\)
\(\Leftrightarrow\frac{x+4}{\left(\sqrt{x+1}-1\right)^2}=1\Leftrightarrow x+4=\left(\sqrt{x+1}-1\right)^2\)
\(\Leftrightarrow x+4=x+2-2\sqrt{x+1}\) (rút gọn vế phải)
\(\Leftrightarrow\sqrt{x+1}=-1\left(\text{vô lí}\right)\)
Vậy pt vô nghiệm
-1; -6
b) ĐK: \(x^2+7x+7\ge0\) (đk xấu quá em ko giải đc;v)
PT \(\Leftrightarrow3x^2+21x+18+2\left(\sqrt{x^2+7x+7}-1\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+2\left(\frac{x^2+7x+6}{\sqrt{x^2+7x+7}+1}\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x+6\right)+\frac{2\left(x+1\right)\left(x+6\right)}{\sqrt{x^2+7x+7}+1}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[3+\frac{1}{\sqrt{x^2+7x+7}+1}\right]=0\)
Hiển nhiên cái ngoặc vuông > 0 nên vô nghiệm suy ra x = -1 (TM) hoặc x = -6 (TM)
Vậy....
P/s: Cũng may nghiệm đẹp chứ chứ nghiệm xấu thì tiêu rồi:(
a) We have \(3\le x\le5\)
Suppose \(\sqrt{5-x}=a\left(a\ge0\right);\sqrt{x-3}=b\left(b\ge0\right)\)
We imidiately have \(a+b=\sqrt{2}\)
On the other hand, we have \(a^2-b^2=\left(\sqrt{5-x}\right)^2-\left(\sqrt{x-3}\right)^2=\left(5-x\right)-\left(x-3\right)=8\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=8\) \(\Leftrightarrow\sqrt{2}\left(a-b\right)=8\) \(\Leftrightarrow a-b=4\sqrt{2}\)
And now we simply have \(\left\{{}\begin{matrix}a+b=\sqrt{2}\\a-b=4\sqrt{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\sqrt{2}}{2}\left(take\right)\\b=\dfrac{3\sqrt{2}}{2}\left(take\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{5-x}=\dfrac{5\sqrt{2}}{2}\\\sqrt{x-3}=\dfrac{3\sqrt{2}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5-x=\dfrac{25}{2}\\x-3=\dfrac{9}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{15}{2}\left(eliminate\right)\\x=\dfrac{15}{2}\left(take\right)\end{matrix}\right.\)
Therefore, this equation have the root \(x=\dfrac{15}{2}\)
b) We have \(x\ge2\)
\(\sqrt{x^2-4}=2\sqrt{x-2}\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}-2\sqrt{x-2}=0\) \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=4\end{matrix}\right.\Leftrightarrow x=2\left(take\right)\)
So this equation has the root \(x=2\)
c) We have \(x\ge-\dfrac{4}{3}\). Suppose \(2x+3=A\left(A\ge0\right);3x+4=B\left(B\ge0\right)\). Notice that \(A+B=2x+3+3x+4=5x+7\). Thus, we can rewrite the equation as below:
\(\sqrt{A+B}=\sqrt{A}+\sqrt{B}\) (1)
Remember that \(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\left(A,B\ge0\right)\). We can prove it easily by squaring each side of this inequality:
\(\sqrt{A}+\sqrt{B}\ge\sqrt{A+B}\Leftrightarrow\left(\sqrt{A}+\sqrt{B}\right)^2\ge\left(\sqrt{A+B}\right)^2\)\(\Leftrightarrow A+B+2\sqrt{AB}\ge A+B\Leftrightarrow2\sqrt{AB}\ge0\Leftrightarrow\sqrt{AB}\ge0\). This is always true when \(A,B\ge0\). "=" happens when one of the expression A, B is equal to 0.
Therefore, (1) happens when \(\left[{}\begin{matrix}A=0\\B=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(eliminate\right)\\x=-\dfrac{4}{3}\left(take\right)\end{matrix}\right.\)
Thus, this equation has the root \(x=-\dfrac{4}{3}\)
d) We have \(x\ge-\dfrac{1}{3}\)
We can see that \(x=-\dfrac{1}{11}\). So, \(\sqrt{5x+7}=\dfrac{6\sqrt{22}}{11}\); o\(\sqrt{x+3}=\dfrac{4\sqrt{22}}{11}\) and \(\sqrt{3x+1}=\dfrac{2\sqrt{22}}{11}\). Therefore, we can rewrite the equation as below:
\(\left(\sqrt{5x+7}-\dfrac{6\sqrt{22}}{11}\right)-\left(\sqrt{x+3}-\dfrac{4\sqrt{22}}{11}\right)-\left(\sqrt{3x+1}-\dfrac{2\sqrt{22}}{11}\right)=0\)Now you multiply and devide each of the terms by the conjugate expression. The first term will has \(5x+\dfrac{5}{11}=5\left(x+\dfrac{1}{11}\right)\) as the numerator, the second term's numerator will be \(x+\dfrac{1}{11}\), and the final term has \(3x+\dfrac{3}{11}=3\left(x+\dfrac{1}{11}\right)\) as the numerator. And now you can see the commom factor \(x+\dfrac{1}{11}\)
\(\left(\sqrt{A}-B=\dfrac{\left(\sqrt{A}-B\right)\left(\sqrt{A}+B\right)}{\sqrt{A}+B}=\dfrac{A-B^2}{\sqrt{A}-B}\right)\)