`#040911`

Các số kế tiếp nhau cách nhau 100 đơn vị

`\Rightarrow` Số thay cho dấu ? là: `43652 + 100 = 43752`

43 752

1.B    2.A

1c

2b

\(BC^2=AB^2+AC^2\Rightarrow AC^2=BC^2-AB^2=2500-900=1600\left(Pitago\right)\)

\(\Rightarrow AC=40\left(cm\right)\)

\(AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{30.40}{50}=24\left(cm\right)\)

\(AB^2=AH^2+HB^2\Rightarrow HB^2=AB^2-AH^2=900-576=324\left(Pitago\right)\)

\(\Rightarrow HB=18\left(cm\right)\)

\(HC=BC-HB=50-18=32\left(cm\right)\)

-2³+0,5.x=1,5

⇒ -8+0,5.x=1,5

⇒ 0,5.x=1,5-(-8)

⇒ 0,5.x=9,5

⇒ x=9,5:0,5

⇒ x=19

-2^{3 _{+ 0,5x = 1,5}}

^{_{-8 + 0,5x=1,5}}

^{_{0,5x=1,5- -8}}

^_0,5x=9,5

^{_{x= 9,5: 0,5}}

^_x=19

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

Xét hình thang ABCD ta có :

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

mà \(\left\{{}\begin{matrix}\widehat{B}+\widehat{D}=180^o\left(đề.bài\right)\\\widehat{B}+\widehat{A}=180^o\left(t/c.hình.thang\right)\end{matrix}\right.\)

\(\Rightarrow\widehat{C}=\widehat{D}\)

⇒ ABCD là hình thang cân (dpcm)

Ta có : AB // CD ⇒ \(\widehat{B}\) + \(\widehat{C}\) = 180^o mà \(\widehat{B}+\widehat{D}=\) 180^o ⇒ \(\widehat{D}=\widehat{C}\)

Vì AB // CD; \(\widehat{D}=\widehat{C}\) vậy ABCD là hình thang cân

a,x.(3\4+2\5)=1

x.20\23=1

x=1:20\23

x=20\23

b,x-9\11=0 hoặc x-25\31=0

x=9\11                x=25\31

c,x-3\7.9\14=7\3

x-2\3=7\3

x=7\3+2\3

x=9\3

x=3

a)Đổi: nửa giờ = 30 phút

 Xe thứ nhất tới B khi: 11 + 1 = 12 (giờ)
 Xe thứ ba tới B khi: 11 giờ - 30 phút = 10 giờ 30 phút 

a)Đổi: nửa giờ = 30 phút 

Xe thứ nhất tới B khi: 11 + 1 = 12 (giờ)

 Xe thứ ba tới B khi: 11 giờ - 30 phút = 10 giờ 30 phút

a) \(\dfrac{x}{12}=\dfrac{6}{9}\)

\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)

b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)

\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)

\(\Rightarrow x=\dfrac{2}{7}\)

c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)

\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)

\(\Rightarrow2x+\dfrac{12}{5}=x\)

\(\Rightarrow2x-x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}\)

\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)