mn giúp mink 

mn giúp mink nha

=>(-4) x y -5= -9

=>(-4) x y= -4

=> y=1

1. Đề lỗi

2.

Đường tròn (C) tâm \(I\left(1;-1\right)\) bán kính \(R=\sqrt{1^2+\left(-1\right)^2-\left(-7\right)}=3\)

a.

\(d\left(I;D\right)=\dfrac{\left|1-1-4\right|}{\sqrt{1^2+1^2}}=2\sqrt{2}< R\)

\(\Rightarrow D\) cắt (C) tại 2 điểm phân biệt

b.

Gọi H là trung điểm MN \(\Rightarrow IH\perp MN\Rightarrow IH=d\left(I;D\right)=2\sqrt{2}\)

ÁP dụng định lý Pitago trong tam giác vuông IHM:

\(HM=\sqrt{IM^2-IH^2}=\sqrt{R^2-IH^2}=\sqrt{9-8}=1\)

\(\Rightarrow MN=2MH=2\)

\(S_{IMN}=\dfrac{1}{2}IH.MN=2\sqrt{2}\)

3.

Đường tròn (C) tâm \(I\left(2;3\right)\) bán kính \(R=\sqrt{2}\)

Đường còn (C') tâm \(I'\left(1;2\right)\) bán kính \(R'=2\sqrt{2}\)

Gọi tiếp tuyến chung của (C) và (C') là (d) có pt: \(ax+by+c=0\) với \(a^2+b^2\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}d\left(I;\left(d\right)\right)=R\\d\left(I';\left(d\right)\right)=R'\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\left|2a+3b+c\right|}{\sqrt{a^2+b^2}}=\sqrt{2}\left(1\right)\\\dfrac{\left|a+2b+c\right|}{\sqrt{a^2+b^2}}=2\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\left|a+2b+c\right|=2\left|2a+3b+c\right|\)

\(\Rightarrow\left[{}\begin{matrix}4a+6b+2c=a+2b+c\\4a+6b+2c=-a-2b-c\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3a+4b+c=0\\5a+8b+3c=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}c=-3a-4b\\c=-\dfrac{5a+8b}{3}\end{matrix}\right.\)

Thế vào (1):

\(\Rightarrow\left[{}\begin{matrix}\dfrac{\left|2a+3b-3a-4b\right|}{\sqrt{a^2+b^2}}=\sqrt{2}\\\dfrac{\left|2a+3b-\dfrac{5a+8b}{3}\right|}{\sqrt{a^2+b^2}}=\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left|a+b\right|=\sqrt{2\left(a^2+b^2\right)}\\\left|a+b\right|=3\sqrt{2\left(a^2+b^2\right)}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a^2+2ab+b^2=2a^2+2b^2\\a^2+2ab+b^2=18a^2+18b^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2=0\\17a^2-2ab+17b^2=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow a=b\) \(\Rightarrow c=-3a-4b=-7a\)

Thế vào pt (d):

\(ax+ay-7a=0\Leftrightarrow x+y-7=0\)

4.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

\(4\left(x+1\right)^2< \left(x+10\right)\left(1-\sqrt{3+2x}\right)^2\)

\(\Leftrightarrow4\left(x+1\right)^2< \left(x+10\right)\left(\dfrac{-2-2x}{1+\sqrt{3+2x}}\right)^2\)

\(\Leftrightarrow4\left(x+1\right)^2< \dfrac{\left(x+10\right)4\left(x+1\right)^2}{\left(1+\sqrt{3+2x}\right)^2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{x+10}{\left(1+\sqrt{3+2x}\right)^2}>1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x+10>1+3+2x+2\sqrt{3+2x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\6-x>2\sqrt{3+2x}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\6-x>0\\\left(6-x\right)^2>4\left(3+2x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x< 6\\x^2-20x+24>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x< 10-2\sqrt{19}\end{matrix}\right.\)

Kết hợp ĐKXĐ ta được nghiệm của BPT là: 

\(\left[{}\begin{matrix}-\dfrac{3}{2}\le x< -1\\-1< x< 10-2\sqrt{19}\end{matrix}\right.\)

C

Chọn C

MN = \(\sqrt{\left(-3-0\right)^2+\left(2+2\right)^2}=5\)

Phép vị tự tâm I biến M và N thành M' và N' với tỉ số vị tự là \(-\dfrac{4}{3}\)

⇒ \(\overrightarrow{M'N'}=-\dfrac{4}{3}\overrightarrow{MN}\)

⇒ M'N' = \(\dfrac{4}{3}.MN=\dfrac{4}{3}.5=\dfrac{20}{3}\) (đơn vị độ dài)

Câu 3:

\(n_{CO_2}=\dfrac{0,44}{44}=0,01\left(mol\right)\)

\(n_{H_2O}=\dfrac{0,18}{18}=0,01\left(mol\right)\)

Bảo toàn C: n_C(A) = 0,01 (mol)

Bảo toàn H: n_C(A) = 2.0,01 = 0,02 (mol)

=> \(n_O=\dfrac{0,3-0,01.12-0,02.1}{16}=0,01\left(mol\right)\)

n_C : n_H : n_O = 0,01 : 0,02 : 0,01 = 1:2:1

=> CTHH: (CH₂O)_n

Có\(n_{O_2}=\dfrac{0,32}{32}=0,01\left(mol\right)=>M_A=\dfrac{0,3}{0,01}=30\left(g/mol\right)\)

=> n = 1

=> CTHH: CH₂O

Câu 4:

\(n_{NO_2}=\dfrac{5,152}{22,4}=0,23\left(mol\right)\)

PTHH: Cu + 4HNO₃ --> Cu(NO₃)₂ + 2NO₂ + 2H₂O

_____a---------------------------------->2a

Fe + 6HNO₃ --> Fe(NO₃)₃ + 3NO₂ + 3H₂O

b---------------------------------->3b

=> \(\left\{{}\begin{matrix}64a+56b=5,36\\2a+3b=0,23\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,04\\b=0,05\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%Cu=\dfrac{0,04.64}{5,36}.100\%=47,76\%\\\%Fe=\dfrac{0,05.56}{5,36}.100\%=52,24\%\end{matrix}\right.\)

Câu 5:

$\frac{20}{\sqrt{5}}=\frac{20\sqrt{5}}{5}=4\sqrt{5}$

Câu 6:

\(\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{3}{\sqrt{5}-\sqrt{2}}=3.\frac{\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}=3.\frac{2\sqrt{5}}{5-2}=2\sqrt{5}\)

Câu 7:

1. ĐKXĐ: $x\neq 1; x\geq 0$

\(A=\left[\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}+1}+1\right]:\left[\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}-1\right]=(\sqrt{x}+1):(\sqrt{x}-1)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

2.

\(A< 1\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{2}{\sqrt{x}-1}<0\)

\(\Leftrightarrow \sqrt{x}-1<0\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ suy ra $0\leq x< 1$