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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) \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right):\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x^2-25\right)+25x+250+x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{10x^2-250+25x+250-x^3}{5x\left(x+5\right)}\right).\dfrac{7}{3\left(x+5\right)}\)
\(C=\dfrac{x\left(x+2.x.5+25\right)}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x+5}{5}.\dfrac{7}{3\left(x+5\right)}=\dfrac{7}{15}\)
Câu 1: D. \(\frac{1}{2}-4x=0\)
Câu 2: C. 2x - 1 = x
Câu 3: D. S = {-9}
# Chúc bạn học tốt #
\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .
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 \(S=\left\{\dfrac{-1}{2};\dfrac{9\pm\sqrt{42}}{3}\right\}\)
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
L
Chọn B
B