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a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
Mik đăng câu hỏi mà ko thấy ai trả lời hết, với lại h mik giải được rồi nên đăng lên có ai tìm bài này thì có đáp án ha ( mấy CTV đừng hiểu lầm nhé)
a) \(x^2-13x+50=4\sqrt{x-3}\)
ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow x^2-13x+50-4\sqrt{x-3}=0\)
\(\Leftrightarrow x^2-14x+x+49-3-+4-4\sqrt{x-3}=0\)
\(\Leftrightarrow(x^2-14x+49)+(x-3-4\sqrt{x-3}+4)=0\)
\(\Leftrightarrow\left(x-7\right)^2+\left(\sqrt{x-3}-2\right)^2=0\)
\(\Leftrightarrow\left(x-7\right)^2=\left(\sqrt{x-3}-2\right)^2\)
\(\Leftrightarrow x-7=-\sqrt{x-3}+2\)
\(\Leftrightarrow x-9=-\sqrt{x-3}\)
\(\Leftrightarrow x^2-18x+81=x-3\)
\(\Leftrightarrow x^2-19x+84=0\)
\(\Leftrightarrow\left(x+12\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy \(x\in\left\{7;12\right\}\)
\(b)\dfrac{4x}{x^2-5x+6}+\dfrac{3x}{x^2-7x+6}=6\)
ĐKXĐ: \(x\ne1,2,3,6\)
Đặt \(t=x^2-6x+6\)
pt \(\Leftrightarrow\dfrac{4x}{t+x}+\dfrac{3x}{t-x}=6\)
\(\Leftrightarrow\dfrac{4x\left(t-x\right)+3x\left(t+x\right)}{\left(t+x\right)\left(t-x\right)}=6\)
\(\Leftrightarrow\dfrac{7tx-x^2}{t^2-x^2}=6\)
\(\Leftrightarrow7tx-x^2=6t^2-6x^2\)
\(\Leftrightarrow-6t^2+7xt+5x^2=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)\left(t-\dfrac{5}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\t-\dfrac{5}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x^2-6x+6-\dfrac{5}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x^2-6x+\dfrac{13}{3}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{9+\sqrt{42}}{3}\\x=\dfrac{9-\sqrt{42}}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{-1}{2};\dfrac{9\pm\sqrt{42}}{3}\right\}\)
1/
a/ \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=3+\sqrt{3}-3-2\sqrt{3}=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
b/ \(\sqrt{12}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
3/ \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(=\left(\dfrac{2\left(x-5\right)}{x}+\dfrac{5\left(x+10\right)}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\left(\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{10x^2-250+25x+250+x^3}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{7\left(x+5\right)^2}{5\left(x+5\right)\cdot3\left(x+5\right)}=\dfrac{7}{15}\)
3) \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right):\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x^2-25\right)+25x+250+x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{10x^2-250+25x+250-x^3}{5x\left(x+5\right)}\right).\dfrac{7}{3\left(x+5\right)}\)
\(C=\dfrac{x\left(x+2.x.5+25\right)}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x+5}{5}.\dfrac{7}{3\left(x+5\right)}=\dfrac{7}{15}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
= \(\dfrac{\left(x+\sqrt{x}+1\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)
= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{x+2}{\sqrt{x}-1}\)
Câu 1:
Tìm max:
Áp dụng BĐT Bunhiacopxky ta có:
\(y^2=(3\sqrt{x-1}+4\sqrt{5-x})^2\leq (3^2+4^2)(x-1+5-x)\)
\(\Rightarrow y^2\leq 100\Rightarrow y\leq 10\)
Vậy \(y_{\max}=10\)
Dấu đẳng thức xảy ra khi \(\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{4}\Leftrightarrow x=\frac{61}{25}\)
Tìm min:
Ta có bổ đề sau: Với $a,b\geq 0$ thì \(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
Chứng minh:
\(\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}\)
\(\Leftrightarrow (\sqrt{a}+\sqrt{b})^2\geq a+b\)
\(\Leftrightarrow \sqrt{ab}\geq 0\) (luôn đúng).
Dấu "=" xảy ra khi $ab=0$
--------------------
Áp dụng bổ đề trên vào bài toán ta có:
\(\sqrt{x-1}+\sqrt{5-x}\geq \sqrt{(x-1)+(5-x)}=2\)
\(\sqrt{5-x}\geq 0\)
\(\Rightarrow y=3(\sqrt{x-1}+\sqrt{5-x})+\sqrt{5-x}\geq 3.2+0=6\)
Vậy $y_{\min}=6$
Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-1)(5-x)=0\\ 5-x=0\end{matrix}\right.\Leftrightarrow x=5\)
Bài 2:
\(A=\sqrt{(x-1994)^2}+\sqrt{(x+1995)^2}=|x-1994|+|x+1995|\)
Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:
\(A=|x-1994|+|x+1995|=|1994-x|+|x+1995|\geq |1994-x+x+1995|=3989\)
Vậy \(A_{\min}=3989\)
Đẳng thức xảy ra khi \((1994-x)(x+1995)\geq 0\Leftrightarrow -1995\leq x\leq 1994\)
1) Đk: \(x\ge4\)
\(\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\sqrt{x-3}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}}=\dfrac{7}{\sqrt{x-3}}\)
\(\Leftrightarrow\dfrac{\sqrt{x^2-16}+x-10}{\sqrt{x-3}}=0\)
\(\Leftrightarrow\sqrt{x^2-16}+x-10=0\)
\(\Leftrightarrow\sqrt{x^2-16}=10-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-16=100-20x+x^2\\x\le10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x=116\\x\le10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\left(N\right)\\x\le10\end{matrix}\right.\)
Kl: x= 29/5
2) Đk: \(x\ge-1\)
\(x^2-5x+14=4\sqrt{x+1}\)
\(\Leftrightarrow x^4+25x^2+196-10x^3-140x+28x^2=16x+16\)
\(\Leftrightarrow x^4-10x^3+53x^2-156x+180=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3-7x^2+32x-60\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x^2-4x+20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-4x+20=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=3\left(N\right)\)
Kl: x=3
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