A

BÀi 4:

2: \(\frac{5}{4-\sqrt{14}}+\frac{1}{3-\sqrt7}-\frac{6}{3+\sqrt7}-\frac{\sqrt7-5}{2}\)
\(=\frac{5\left(4+\sqrt{14}\right)}{16-14}+\frac{3+\sqrt7}{\left(3-\sqrt7\right)\left(3+\sqrt7\right)}-\frac{6\left(3-\sqrt7\right)}{\left(3+\sqrt7\right)\left(3-\sqrt7\right)}-\frac{\sqrt7-5}{2}\)

\(=\frac{20+5\sqrt{14}}{2}+\frac{3+\sqrt7}{2}-\frac{18-6\sqrt7}{2}-\frac{\sqrt7-5}{2}=\frac{20+5\sqrt{14}+3+\sqrt7-18+6\sqrt7-\sqrt7+5}{2}=\frac{10+5\sqrt{14}+6\sqrt7}{2}\)

3: \(\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\cdots+\frac{1}{\sqrt{2020}+\sqrt{2021}}\)

\(=\frac{-1+\sqrt2}{\left(1+\sqrt2\right)\left(-1+\sqrt2\right)}+\frac{-\sqrt2+\sqrt3}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+...+\frac{-\sqrt{2020}+\sqrt{2021}}{\left(\sqrt{2020}+\sqrt{2021}\right)\left(\sqrt{2021}-\sqrt{2020}\right)}\)

\(=-1+\sqrt2-\sqrt2+\sqrt3-\cdots-\sqrt{2020}+\sqrt{2021}=\sqrt{2021}-1\)

4: \(\frac{1}{1-\sqrt2}-\frac{1}{\sqrt2-\sqrt3}+\frac{1}{\sqrt3-\sqrt4}-\cdots+\frac{1}{\sqrt{143}-\sqrt{144}}\)

\(=\frac{1\left(1+\sqrt2\right)}{\left(1-\sqrt2\right)\left(1+\sqrt2\right)}-\frac{\left(\sqrt2+\sqrt3\right)}{\left(\sqrt2-\sqrt3\right)\left(\sqrt2+\sqrt3\right)}+\frac{\left(\sqrt3+\sqrt4\right)}{\left(\sqrt3-\sqrt4\right)\left(\sqrt3+\sqrt4\right)}+\cdots+\frac{\sqrt{143}+\sqrt{144}}{\left(\sqrt{143}+\sqrt{144}\right)\left(\sqrt{143}-\sqrt{144}\right)}\)

\(=-1-\sqrt2+\sqrt2+\sqrt3-\sqrt3-\sqrt4+\cdots-\sqrt{143}-\sqrt{144}\)

=-1-12

=-13

5: \(\sqrt2\cdot\left(\sqrt3+1\right)\cdot\sqrt{2-\sqrt3}\)

\(=\left(\sqrt3+1\right)\cdot\sqrt{4-2\sqrt3}\)

\(=\left(\sqrt3+1\right)\cdot\sqrt{\left(\sqrt3-1\right)^2}=\left(\sqrt3+1\right)\left(\sqrt3-1\right)=3-1=2\)

6: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt6\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\left(\sqrt5-\sqrt3\right)=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\cdot\left(16-15\right)=2\)

7: \(\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt3}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt3+1-\sqrt3+1\right)=\frac{2}{\sqrt2}=\sqrt2\)

Bài 5:

1: \(x\cdot\sqrt{27}+\sqrt8=x\cdot\sqrt{18}+\sqrt{12}\)

=>\(x\cdot3\sqrt3-x\cdot3\sqrt2=2\sqrt3-2\sqrt2\)

=>\(3x\left(\sqrt3-\sqrt2\right)=2\left(\sqrt3-\sqrt2\right)\)

=>3x=2

=>\(x=\frac23\)

2: \(2x\cdot\sqrt3+\sqrt{48}=x\cdot\sqrt{75}+\sqrt{243}\)

=>\(x\cdot2\sqrt3+4\sqrt3=x\cdot5\sqrt3+9\sqrt3\)

=>\(x\cdot\left(-3\sqrt3\right)=9\sqrt3-4\sqrt3=5\sqrt3\)

=>\(x=-\frac53\)

4: ĐKXĐ: x>=2

\(\sqrt{49x-98}-\sqrt{9x-18}-\sqrt{16x-32}=\sqrt{4x-4}\)

=>\(7\cdot\sqrt{x-2}-3\cdot\sqrt{x-2}-4\cdot\sqrt{x-2}=2\sqrt{x-1}\)

=>\(2\sqrt{x-1}=0\)

=>x-1=0

=>x=1(loại)

5: ĐKXĐ: \(x^2-2\ge0\)

=>\(x^2\ge2\)

=>\(\left[\begin{array}{l}x\ge\sqrt2\\ x\le-\sqrt2\end{array}\right.\)

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

=>\(\sqrt{x^2-2}=1\)

=>\(x^2-2=1\)

=>\(x^2=3\)

=>\(\left[\begin{array}{l}x=\sqrt3\left(nhận\right)\\ x=-\sqrt3\left(nhận\right)\end{array}\right.\)

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: |A|=3

=>A=-3 hoặc A=3

=>x+1=3x-9 hoặc x+1=-3x+9

=>-2x=-10 hoặc 4x=8

=>x=2(nhận) hoặc x=5(nhận)

Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O (1)

CaO + 2HCl ---> CaCl₂ + H₂O

a. Theo PT₍₁₎: \(n_{CaCO_3}=n_{CO_2}=0,15\left(mol\right)\)

=> \(m_{CaCO_3}=0,15.100=15\left(g\right)\)

=> \(\%_{m_{CaCO_3}}=\dfrac{15}{17,8}.100\%=84,27\%\%\)

\(\%_{m_{CaO}}=100\%-84,27\%=15,73\%\)

b. Ta có: \(m_{CaO}=17,8-15=2,8\left(g\right)\)

=> \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

Theo PT₍₁₎: \(n_{HCl}=2.n_{CaCO_3}=2.0,15=0,3\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{CaO}=2.0,05=0,1\left(mol\right)\)

=> \(n_{HCl_{PỨ}}=0,3+0,1=0,4\left(mol\right)\)

Đổi 200ml = 0,2 lít

=> \(C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\)

\(CD=\sqrt{0,2^2+0,5^2}=\dfrac{\sqrt{29}}{10}\left(cm\right)\)

Xét ΔCED vuông tại E và ΔCAB vuông tại A có

góc C chung

=>ΔCED đồng dạng với ΔCAB

=>DE/AB=CD/CB

=>\(\dfrac{0.5}{AB}=\dfrac{\sqrt{29}}{10}:10=\dfrac{\sqrt{29}}{100}\)

=>\(AB=\dfrac{50}{\sqrt{29}}\)(cm)

=>Ko có câu nào đúng

Ta có : \(\dfrac{56}{24}>1\) do tử > mẫu

\(\dfrac{55}{33}>1\) do tử > mẫu

\(\dfrac{126}{168}< 1\) tử bé hơn mẫu nên ta chỉ xét hai phân số : \(\dfrac{56}{24};\dfrac{55}{33}\)

Ta lại có : \(\dfrac{56}{24}=\dfrac{7}{3}\)

\(\dfrac{55}{33}=\dfrac{5}{3}\)

do 7>5 suy ra : \(\dfrac{56}{24}>\dfrac{55}{33}\)

\(\Rightarrow\dfrac{126}{168}< \dfrac{55}{33}< \dfrac{56}{24}\)

\(\dfrac{56}{24}=\dfrac{7}{3}\) ( cùng chia cả tử và mẫu cho 8 )

\(\dfrac{126}{168}=\dfrac{3}{4}\)( cùng chia cả tử và mẫu cho 42 ) 

\(\dfrac{55}{33}=\dfrac{5}{3}\) ( cùng chia cả tử và mẫu cho 11 )

Đk:\(y^2-2x-5y+6\ge0\)

Pt (1)\(\Leftrightarrow\left(x^2-1\right)-\left(xy-y\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-y\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+2\end{matrix}\right.\)

TH1: Thay x=1 vào pt (2) ta đc: \(3\sqrt{y^2-5y+4}=y+9\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+9\ge0\\9\left(x^2-5y+4\right)=y^2+18y+81\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y\ge-9\\8y^2-63y-45=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{63+3\sqrt{601}}{16}\\y=\dfrac{63-3\sqrt{601}}{16}\end{matrix}\right.\) (tm)

TH2: Thay y=x+2 vào pt (2) ta đc:

\(\left(x-1\right)^2+3\sqrt{\left(x+2\right)^2-2x-5\left(x+2\right)+6}=x+2+9\)

\(\Leftrightarrow x^2-3x-10+3\sqrt{x^2-3x}=0\)

Đặt \(t=\sqrt{x^2-3x}\left(t\ge0\right)\)

Pttt: \(t^2-10+3t=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow2=\sqrt{x^2-3x}\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=6\\y=1\end{matrix}\right.\) (tm)

Vậy \(\left(x;y\right)=\text{​​}\left\{\left(1;\dfrac{63+3\sqrt{601}}{16}\right);\left(1;\dfrac{63-3\sqrt{601}}{16}\right),\left(4;6\right),\left(-1;1\right)\right\}\)

Xét pt đầu:

\(\left(x^2+x-2\right)-y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+2\end{matrix}\right.\)

- Với \(x=1\) thay xuống pt dưới:

\(3\sqrt{y^2-5y+4}=y+9\) \(\left(y\ge-9\right)\)

\(\Leftrightarrow9\left(y^2-5y+4\right)=y^2+18y+81\)

\(\Leftrightarrow8y^2-63y-45=0\)

\(\Rightarrow y=\dfrac{63\pm3\sqrt{601}}{16}\) (thỏa mãn)

- Với \(y=x+2\) thay xuống pt dưới:

\(\left(x-1\right)^2+3\sqrt{x^2-3x}=x+11\) (ĐKXĐ: ....)

\(\Leftrightarrow x^2-3x+3\sqrt{x^2-3x}-10=0\)

Đặt \(\sqrt{x^2-3x}=t\ge0\)

\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-3x}=2\Leftrightarrow x^2-3x-4=0\)

\(\Leftrightarrow...\)