a: \(B=\dfrac{x-2\sqrt{x}}{\sqrt{x}-2}-\dfrac{2x+12\sqrt{x}+18}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}-\dfrac{2\left(x+6\sqrt{x}+9\right)}{\sqrt{x}+3}\)

=căn x-2(căn x+3)

=-căn x-6

b: B+8>0

=>-căn x-6+8>0

=>-căn x+2>0

=>-căn x>-2

=>căn x<2

=>0<=x<4

\(A=\dfrac{\sqrt{2}}{\sqrt{2}+1}+\dfrac{1}{3}\cdot3\sqrt{2}\)

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

ĐKXĐ: x ≥ 2

Phương trình đã cho tương đương:

√(x - 2) + 6√(x - 2) - 2√(x - 2) = 10

⇔ 5√(x - 2) = 10

⇔ √(x - 2) = 2

⇔ x - 2 = 4

⇔ x = 6 (nhận)

Vậy S = {6}

\(A=4\sqrt{2}sinx+1-2sin^2x+2=-2sin^2x+4\sqrt{2}sinx+3\)

Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)

\(A=f\left(t\right)=-2t^2+4\sqrt{2}t+3\)

Xét hàm \(f\left(t\right)\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=-\sqrt{2}\notin\left[-1;1\right]\)

\(f\left(-1\right)=1-4\sqrt{2}\) ; \(f\left(1\right)=1+4\sqrt{2}\)

\(\Rightarrow A_{max}=f\left(1\right)=1+4\sqrt{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=2\end{matrix}\right.\)

Ủa đề bài sai, \(c>a\) chứ sao \(c\le a\) được?

//Em xem lại câu hỏi hồi nãy nhé, lúc nhấn gửi đáp án mới làm được 1 nửa nên chưa đúng đâu

\(2\sqrt{a}-a\sqrt{\dfrac{4}{a}}\)

\(=2\sqrt{a}-a.\dfrac{\sqrt{4}}{\sqrt{a}}\)

\(=2\sqrt{a}-a.\dfrac{2}{\sqrt{a}}\)

\(=2\sqrt{a}-2\sqrt{a}\)

\(=0\)

=2/5(-3/7-5/14)-18/35

=2/5x(-11/14)-18/35

=-22/70-18/35=-29/35

-3/7 . 2/5 + 2/5 . (-5/14) - 18/35

= 2/5 . [ -3/7 + (-5/14) ] - 18/35

= 2/5 . [ -6/14 + (-5/14) ] - 18/35

= 2/5 . -11/14 - 18/35

= -11/35 - 18/35

= -29/35.

\(\dfrac{-3}{7}\cdot\dfrac{2}{5}+\dfrac{2}{5}\cdot\dfrac{-5}{14}-\dfrac{18}{35}=\dfrac{2}{5}\cdot\left(\dfrac{-3}{7}+\dfrac{-5}{14}\right)-\dfrac{18}{35}=\dfrac{2}{5}\cdot\left(-\dfrac{11}{15}\right)-\dfrac{18}{35}=\left(-\dfrac{11}{35}\right)-\dfrac{18}{35}=-\dfrac{29}{35}\)

a,ĐKXĐ:\(x\ge2\)

\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

b,ĐKXĐ:\(x\in R\)

\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(x\ge0\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)

Câu 18:

\(n_{NO}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \text{Đ}\text{ặt}:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\\ Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,15\\27a+56b=5,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,1.27}{5,5}.100\approx49,091\%;\%m_{Fe}\approx50,909\%\\ b,n_{HNO_3}=4.n_{NO}=0,6\left(mol\right)\\ V_{\text{dd}HNO_3}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c) Dung dịch Y là dung dịch nào?