Toán

ĐKXĐ: x ≥ 3

Phương trình tương đương:

7√(x - 3) - 2.2√(x - 3) = 12

⇔ 7√(x - 3) - 4√(x - 3) = 12

⇔ 3√(x - 3) = 12

⇔ √(x - 3) = 12 : 3

⇔ √(x - 3) = 4

⇒ x - 3 = 16

⇔ x = 16 + 3

⇔ x = 19 (nhận)

Vậy S = {19}

\(7\sqrt{x-3}-2\sqrt{4x-12}=12\\ \Leftrightarrow7\sqrt{x-3}-2\sqrt{4\left(x-3\right)}=12\\ \Leftrightarrow7\sqrt{x-3}-4\sqrt{x-3}=12\\ \Leftrightarrow3\sqrt{x-3}=12\\ \Leftrightarrow\sqrt{x-3}=\dfrac{12}{3}\\ \Leftrightarrow\sqrt{x-3}=4\\ \Leftrightarrow\left(\sqrt{x-3}\right)^2=4^2\\ \Leftrightarrow x-3=16\\ \Leftrightarrow x=16+3\\ \Leftrightarrow x=19.\)

a) \(\sqrt{1-x}+\sqrt{x+4}=3\left(Đk:-4\le x\le1\right)\)

\(1-x+x+4+2\sqrt{\left(1-x\right)\left(x+4\right)}=9\)

\(\sqrt{-x^2-3x+4}=2\)

\(-x^2-3x+4=4\)

\(x^2+3x=0\)

\(x\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\left(TM\right)\\x=0\left(TM\right)\end{matrix}\right.\)

b) \(x^2+4x+5=2\sqrt{2x+3}\) \(\left(Đk:x\ge\dfrac{-3}{2}\right)\)

\(\)\(x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\)

\(\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

Vì \(\left(x+1\right)^2;\left(\sqrt{2x+3}-1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+3=1\end{matrix}\right.\)\(\Leftrightarrow x=-1\)(TM)

Đề sai rồi bạn

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x-1\right)^2}{x\left(x^2+5\right)}=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-2x+1}{x\left(x^2+5\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}+\dfrac{1}{x^2}}{x\left(1+\dfrac{5}{x^2}\right)}=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x}\cdot\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}+\dfrac{1}{x^2}}{1+\dfrac{5}{x^2}}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}\dfrac{1}{x}=+\infty\\\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}+\dfrac{1}{x^2}}{1+\dfrac{5}{x^2}}=\dfrac{1}{1}=1>0\end{matrix}\right.\)

17² - 17 + 17.84

= 17.(17 - 1 + 84)

= 17.100

= 1700

\(\widehat{ABC}=90^o+15^o30'=105,5^o\)

\(\widehat{CAB}=90^o-45^o=45^o\)

\(\widehat{ACB}=180^o-105,5^o-45^o=29,5^o\)

Ta có:

\(\dfrac{AC}{sin105,5}=\dfrac{AB}{sin29,5}\)

\(\Rightarrow AC=\dfrac{120.sin105,5}{sin29,5}=234,8\left(m\right)\)

Ta có:

\(sin45=\dfrac{CH}{AC}\)

\(\Rightarrow CH=234,8.sin45=166\left(m\right)\)

\(P=\dfrac{\sqrt{a}}{a\sqrt{a}-3a+3\sqrt{a}+1}\) \(\left(a\ge0\right)\)

   \(=\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)^3-1}\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\left(\sqrt{a}-1\right)^3-1\ge-2\end{matrix}\right.\)

\(\Rightarrow P\le0\)

\(MaxP=0\Leftrightarrow a=0\)

\(AB^2=AC^2+BC^2-2AC.BC.cosACB\)

          \(=700^2+1000^2-2.700.1000.cos120\)

          \(=2190000\)

\(\Rightarrow AB\simeq1480\left(m\right)\)

35 ⋮ x; 130 ⋮ x và x lớn nhất

⇒ x = ƯCLN(35; 130)

Ta có:

35 = 5.7

130 = 2.5.13

⇒ x = ƯCLN(35; 130) = 5