`#3107.101107`

`1.`

`a)`

\(\dfrac{7}{23}-\dfrac{5}{14}+\dfrac{1}{2}-\dfrac{9}{14}+\dfrac{16}{23}\\ =\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\left(\dfrac{5}{14}+\dfrac{9}{14}\right)+\dfrac{1}{2}\\ =\dfrac{23}{23}-\dfrac{14}{14}+\dfrac{1}{2}\\ =1-1+\dfrac{1}{2}\\ =\dfrac{1}{2}\)

`b)`

\(\left(-\dfrac{2}{3}\right)\cdot\dfrac{3}{11}+\left(-\dfrac{16}{9}\right)\cdot\dfrac{3}{11}\\ =\dfrac{3}{11}\cdot\left(-\dfrac{2}{3}-\dfrac{16}{9}\right)\\ =\dfrac{3}{11}\cdot\left(-\dfrac{22}{9}\right)\\ =-\dfrac{2}{3}\)

`2.`

`a)`

\(-\dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{1}{3}\\ \Rightarrow-\dfrac{3}{7}x=\dfrac{1}{6}\\ \Rightarrow x=\dfrac{1}{6}\div\left(-\dfrac{3}{7}\right)\\ \Rightarrow x=-\dfrac{7}{18}\)

`b)`

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

_____

`1.`

`a)`

\(-\dfrac{9}{17}+6,72+\dfrac{-8}{17}+\left(-4,72\right)\\ =\left(-\dfrac{9}{17}-\dfrac{8}{17}\right)+\left(6,72-4,72\right)\\ =-\dfrac{17}{17}+2\\ =-1+2=1\)

`b)`

\(2\dfrac{1}{5}-\left(-\dfrac{3}{4}+\dfrac{1}{5}\right)\\ =\dfrac{11}{5}+\dfrac{3}{4}-\dfrac{1}{5}\\ =\left(\dfrac{11}{5}-\dfrac{1}{5}\right)+\dfrac{3}{4}\\ =\dfrac{10}{5}-\dfrac{3}{4}\\ =2-\dfrac{3}{4}\\ =\dfrac{5}{4}\)

`c)`

\(\left(-\dfrac{1}{3}\right)^3-\dfrac{3}{8}\div \left(\dfrac{1}{2}\right)^3-\dfrac{5}{2}\cdot\left(-2\right)\\ =\left(-\dfrac{1}{27}\right)-\dfrac{3}{8}\div\dfrac{1}{8}-\left(-5\right)\\ =-\dfrac{1}{27}-3+5\\ =-\dfrac{1}{27}+2\\ =\dfrac{53}{27}\)

`d)`

\(4,1\cdot\dfrac{-5}{12}-6,2+4,1\cdot\dfrac{-7}{12}\\ =4,1\cdot\left(-\dfrac{5}{12}-\dfrac{7}{12}\right)-6,2\\ =4,1\cdot\left(-\dfrac{12}{12}\right)-6,2\\ 4,1\cdot\left(-1\right)-6,2\\ =-4,1-6,2\\ =-10,3\)

`2.`

`a)`

\(x+\left(-\dfrac{1}{9}\right)=-\dfrac{7}{6}\\ \Rightarrow x=-\dfrac{7}{6}-\left(-\dfrac{1}{9}\right)\\ \Rightarrow x=-\dfrac{7}{6}+\dfrac{1}{9}\\ \Rightarrow x=-\dfrac{19}{18}\)

`b)`

\(\left(x-\dfrac{4}{7}\right)\div\dfrac{-5}{3}=0,2\\ \Rightarrow x-\dfrac{4}{7}=0,2\cdot\left(-\dfrac{5}{3}\right)\\ \Rightarrow x-\dfrac{4}{7}=-\dfrac{1}{3}\\ \Rightarrow x=-\dfrac{1}{3}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{5}{21}\)

`c)`

\(\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{8}\\ \Rightarrow\left(2x-\dfrac{1}{3}\right)^3=\left(\dfrac{1}{2}\right)^3\\ \Rightarrow2x-\dfrac{1}{3}=\dfrac{1}{2}\\ \Rightarrow2x=\dfrac{1}{2}+\dfrac{1}{3}\\ \Rightarrow2x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}\div2\\ \Rightarrow x=\dfrac{5}{12}\)

____

`13.`

`a)`

Ta có: \(\dfrac{139}{303}>\dfrac{138}{303}=\dfrac{46}{101}\)

Mà \(\dfrac{35}{101}< \dfrac{46}{101}\)

\(\Rightarrow\dfrac{139}{303}>\dfrac{35}{101}\)

`b)`

Ta có:

\(\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^{8-2}=\left(\dfrac{1}{2}\right)^6\)

\(\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3=\left(\dfrac{1}{2}\right)^{3+3}=\left(\dfrac{1}{2}\right)^6\)

Vì \(\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^6\\ \Rightarrow\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3\)

`14.`

`a)`

\(-\dfrac{11}{24}+\dfrac{3}{4}-\dfrac{13}{24}\\ =\left(-\dfrac{11}{24}-\dfrac{13}{24}\right)+\dfrac{3}{4}\\ =-\dfrac{24}{24}+\dfrac{3}{4}\\ =-1+\dfrac{3}{4}\\ =-\dfrac{1}{4}\)

`b)`

\(-\dfrac{5}{9}-\left(\dfrac{8}{15}+\dfrac{4}{9}\right)+\dfrac{7}{15}\\ =-\dfrac{5}{9}-\dfrac{8}{15}-\dfrac{4}{9}+\dfrac{7}{15}\\ =\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)-\left(\dfrac{8}{15}+\dfrac{7}{15}\right)\\ =-\dfrac{9}{9}-\dfrac{15}{15}\\ =-1-1=-2\)

`c)`

\(\dfrac{5}{9}\div2,4-\dfrac{41}{9}\div2,4\\ =\dfrac{5}{9}\div\dfrac{12}{5}-\dfrac{41}{9}\div\dfrac{12}{5}\\ =\dfrac{5}{9}\cdot\dfrac{5}{12}-\dfrac{41}{9}\cdot\dfrac{5}{12}\\ =\dfrac{5}{12}\cdot\left(\dfrac{5}{9}-\dfrac{41}{9}\right)\\ =\dfrac{5}{12}\cdot\left(-\dfrac{36}{9}\right)\\ =\dfrac{5}{12}\cdot\left(-4\right)\\ =-\dfrac{5}{3}\)

`a)`

`b)`

`c)`

a, \(P\left(x\right)=3x^3+2x^3-2x+7-x^2-x=5x^3-3x+7-x^2\)

\(Q\left(x\right)=-3x^3+x-14-2x-x^2-1=-3x^3-x-x^2-15\)

b, \(M\left(x\right)=5x^3-3x+7-x^2-3x^3-x-x^2-15=2x^3-2x^2-4x-8\)

\(N\left(x\right)=5x^3-3x+7-x^2+3x^3+x+x^2+15=8x^3-2x+22\)

c, \(P\left(x\right)=-Q\left(x\right)\Leftrightarrow5x^3-3x+7-x^2=3x^3+x+x^2+15\)

\(\Leftrightarrow2x^3-2x^2-4x-8=0\)

a) \(P\left(x\right)=3x^3+2x^3-2x+7-x^2-x\\ =\left(3x^3+2x^3\right)-x^2+\left(-2x-x\right)+7\\ =5x^3-x^2-3x+7\)

\(Q\left(x\right)=-3x^3+x-14-2x-x^2-1\\ =-3x^3-x^2+\left(x-2x\right)+\left(-14-1\right)\\ =-3x^3-x^2-x-15\)

b) \(M\left(x\right)=5x^3-x^2-3x+7+\left(-3x^3-x^2-x-15\right)\\ =\left(5x^3-3x^3\right)+\left(-x^2-x^2\right)+\left(-3x-x\right)+\left(7-15\right)\\ =2x^3-2x^2-4x-8\)

 \(N\left(x\right)=5x^3-x^2-3x+7-\left(-3x^3-x^2-x-15\right)\\ =5x^3-x^2-3x+7+3x^3+x^2+x+15\\ =\left(5x^3+3x^3\right)+\left(x^2-x^2\right)+\left(x-3x\right)+\left(15+7\right)\\ =8x^3-2x+22\)

c) \(P\left(x\right)=-Q\left(x\right)\Rightarrow P\left(x\right)+Q\left(x\right)=0\\ \Rightarrow M\left(x\right)=0\Rightarrow2x^3-2x^2-4x-8=0\\ \Rightarrow x^3-x^2-2x-4=0\)

Bạn xem lại đề nhé

Số số hạng của dãy số là:

\(\left(198-2\right):2+1=99\) (số)

Tổng các số hạng trong dãy là:

\(\left(198+2\right)\times99:2=9900\)

Trung bình cộng của các số trong dãy là:

\(9900:99=100\)

Đáp số: 100

Các số chẵn từ 2 đến 198 là: 2; 4; 6; 8; ....; 196; 198

Dãy trên là dãy cách đều 

Do đó TBC dãy trên là:

  (198 + 2) : 2 = 100

2x² - 18x + 6x -6 = 16 + 25

2x² - 12x -47 =0

\(x=\pm\dfrac{\sqrt{130}+6}{2}\)

Hằng đẳng thức: \(a^2+2ab+b^2=\left(a+b\right)^2\)

Cách chứng minh: \(VT=\left(a^2+ab\right)+\left(ab+b^2\right)=a\left(a+b\right)+b\left(a+b\right)\\ =\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2=VP\)

Áp dụng:

Kiểu đề 1: \(2x\left(x-9\right)+3\left(2x\right)-6=4^2+5^2\\ \Rightarrow2x^2-18x+6x-6=16+25\\ \Rightarrow2x^2-12x-47=0\\ \Rightarrow x^2-6x-\dfrac{47}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{47}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{65}{2}=\left(\dfrac{\pm\sqrt{130}}{2}\right)^2\\\)

\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{130}}{2}\\x-3=\dfrac{-\sqrt{130}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{130}}{2}\\x=\dfrac{6-\sqrt{130}}{2}\end{matrix}\right.\)

Kiểu đề 2: \(2x\left(x-9\right)+3\left(2x-6\right)=4^2+5^2\\ \Rightarrow2x^2-18x+6x-18=16+25\\ \Rightarrow2x^2-12x-59=0\\ \Rightarrow x^2-6x-\dfrac{59}{2}=0\\ \Rightarrow\left(x^2-2.x.3+3^2\right)-9-\dfrac{59}{2}=0\\ \Rightarrow\left(x-3\right)^2=\dfrac{77}{2}=\left(\dfrac{\pm\sqrt{154}}{2}\right)^2\\ \)

\(\Rightarrow\left[{}\begin{matrix}x-3=\dfrac{\sqrt{154}}{2}\\x-3=\dfrac{-\sqrt{154}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6+\sqrt{154}}{2}\\x=\dfrac{6-\sqrt{154}}{2}\end{matrix}\right.\)

làm phần c thôi nhé

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

Nhận xét:

\(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0,\forall x,y\)

Dấu \("="\) xảy ra khi:

\(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

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

\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Xét tam giác ABC vuông tại A. Đặt \(\widehat{B}=a\left(0^o< a< 90^o\right)\) 

Khi đó ta có \(\tan a=\dfrac{\sin a}{\cos a}=\dfrac{AC}{AB}< 1\) (vì \(\cos a>\sin a\))

\(\Rightarrow AC< AB\)

\(\Rightarrow\widehat{B}< \widehat{C}\) (quan hệ giữa góc và cạnh đối diện trong tam giác)

Lại có \(\widehat{B}+\widehat{C}=90^o>\widehat{B}+\widehat{B}=2\widehat{B}\)  nên \(\widehat{B}=a< 45^o\).

Ta có đpcm.

Câu 1: Đại dịch Covid-19 gây ra hậu quả khó lường như: số người chết trên toàn thế giới dưới 2 tỷ người, giản cách xã hội rất lâu,những chuyến bay sẽ bị hủy bỏ.

Cửa hàng đã bán tất cả số chai dầu là:

10 x 6 = 60 (chai)

ĐS: 60 chai 

                                             Bài giải

                      Cửa hàng đó bán được số chai dầu là :

                                        10 x 6 = 60 (chai)

                                                   Đáp số : 60 chai dầu .

 Chọn hệ trục tọa độ Mxyz (M là gốc tọa độ) sao cho Mx trùng với tia MB, My trùng với tia MA và Mz cùng phương với BB' sao cho \(\overrightarrow{BB'}\) hướng theo chiều dương của Mz. 

 Gọi chiều cao lăng trụ là \(h>0\)

 Khi đó \(B\left(a;0;0\right)\), \(C'\left(-a;0;h\right)\), \(A'\left(0;a\sqrt{3};h\right)\)

 Ta có \(\overrightarrow{MC'}=\left(-a;0;h\right),\overrightarrow{BA'}=\left(-a;a\sqrt{3};h\right)\)

\(\Rightarrow\left[\overrightarrow{MC'},\overrightarrow{BA'}\right]=\left(-ah\sqrt{3};0;a^2\sqrt{3}\right)\)

\(\Rightarrow\left|\left[\overrightarrow{MC'},\overrightarrow{BA'}\right]\right|=\sqrt{\left(-ah\sqrt{3}\right)^2+\left(a^2\sqrt{3}\right)^2}=a\sqrt{3h^2+3a^2}\)

Lại có \(\overrightarrow{MB}=\left(a;0;0\right)\)

\(\Rightarrow\left[\overrightarrow{MC'},\overrightarrow{BA'}\right].\overrightarrow{MB}=-a^2h\sqrt{3}\)

\(\Rightarrow d\left(MC',BA'\right)=\dfrac{\left|\left[\overrightarrow{MC'},\overrightarrow{BA'}\right].\overrightarrow{MB}\right|}{\left|\left[\overrightarrow{MC'},\overrightarrow{BA'}\right]\right|}\) \(=\dfrac{a^2h\sqrt{3}}{a\sqrt{3a^2+3h^2}}=\dfrac{ah}{\sqrt{a^2+h^2}}\)

Theo đề bài, ta có: \(\dfrac{ah}{\sqrt{a^2+h^2}}=\dfrac{a}{2}\) 

\(\Leftrightarrow\dfrac{h}{\sqrt{a^2+h^2}}=\dfrac{1}{2}\)

\(\Leftrightarrow2h=\sqrt{a^2+h^2}\) 

\(\Leftrightarrow4h^2=a^2+h^2\)

\(\Leftrightarrow3h^2=a^2\)

\(\Leftrightarrow h=\dfrac{a}{\sqrt{3}}\)

\(\Rightarrow V=S_đ.h=\dfrac{\left(2a\right)^2\sqrt{3}}{4}.\dfrac{a}{\sqrt{3}}=a^3\)

Vậy thể tích lăng trụ bằng \(a^3\)