Toán

g: \(\left(\dfrac{8}{5}+\dfrac{2}{5}\right)\cdot\dfrac{5}{7}+\left(\dfrac{6}{5}+\dfrac{9}{5}\right)\cdot\dfrac{5}{7}\)

\(=\dfrac{10}{5}\cdot\dfrac{5}{7}+\dfrac{15}{5}\cdot\dfrac{5}{7}\)

\(=2\cdot\dfrac{5}{7}+3\cdot\dfrac{5}{7}=5\cdot\dfrac{5}{7}=\dfrac{25}{7}\)

h: \(\dfrac{4}{7}\cdot\dfrac{3}{5}\cdot\dfrac{7}{4}\cdot\left(-20\right)\cdot\dfrac{-11}{12}\)

\(=\dfrac{4}{7}\cdot\dfrac{7}{4}\cdot\dfrac{3}{5}\cdot20\cdot\dfrac{11}{12}\)

\(=12\cdot\dfrac{11}{12}=11\)

d: \(\dfrac{7}{13}\cdot\dfrac{5}{19}+\dfrac{7}{19}\cdot\dfrac{8}{13}-3\cdot\dfrac{7}{19}\)

\(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}\right)-3\cdot\dfrac{7}{19}\)

\(=\dfrac{7}{19}-3\cdot\dfrac{7}{19}=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)

e: \(\dfrac{-9}{16}\cdot\dfrac{13}{3}-\left(-\dfrac{3}{4}\right)\cdot\dfrac{19}{3}\)

\(=-3\cdot\dfrac{13}{16}+\dfrac{3}{4}\cdot\dfrac{19}{3}\)

\(=-\dfrac{39}{16}+\dfrac{19}{4}=\dfrac{-39+76}{16}=\dfrac{37}{16}\)

f: \(\dfrac{4}{7}\cdot\dfrac{3}{13}-\dfrac{-2}{7}\cdot\dfrac{3}{13}+\dfrac{1}{7}\cdot\dfrac{-3}{13}\)

\(=\dfrac{4}{7}\cdot\dfrac{3}{13}+\dfrac{2}{7}\cdot\dfrac{3}{13}-\dfrac{1}{7}\cdot\dfrac{3}{13}\)

\(=\dfrac{3}{13}\left(\dfrac{4}{7}+\dfrac{2}{7}-\dfrac{1}{7}\right)=\dfrac{3}{13}\cdot\dfrac{5}{7}=\dfrac{15}{91}\)

g: \(\dfrac{-1}{28}:\dfrac{1}{14}+\dfrac{3}{28}:\dfrac{3}{14}\)

\(=\dfrac{-1}{28}\cdot14+\dfrac{3}{28}\cdot\dfrac{14}{3}\)

\(=-\dfrac{1}{2}+\dfrac{1}{2}=0\)

a: \(\dfrac{5}{23}\cdot\dfrac{17}{26}+\dfrac{5}{23}\cdot\dfrac{9}{26}\)

\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)

\(=\dfrac{5}{23}\cdot\dfrac{26}{26}=\dfrac{5}{23}\)

b: \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)

\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)

\(=1-\dfrac{29}{15}=-\dfrac{14}{15}\)

c: \(\dfrac{5}{8}\cdot\left(-56\right)\cdot\dfrac{5}{7}\cdot\left(-4\right)\)

\(=56\cdot\dfrac{5}{8}\cdot\dfrac{5}{7}\cdot4\)

\(=25\cdot4=100\)

d: \(\dfrac{1}{3}\cdot\dfrac{2}{5}+\dfrac{-2}{6}\cdot\dfrac{2}{5}+\dfrac{-3}{9}\cdot\dfrac{2}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{2}{5}-\dfrac{1}{3}\cdot\dfrac{2}{5}-\dfrac{1}{3}\cdot\dfrac{2}{5}\)

\(=-\dfrac{1}{3}\cdot\dfrac{2}{5}=-\dfrac{2}{15}\)

e: \(\dfrac{-1}{3}\cdot\dfrac{141}{17}-\dfrac{39}{3}\cdot\dfrac{-1}{17}\)

\(=\dfrac{-1}{17}\cdot\dfrac{141}{3}-13\cdot\dfrac{-1}{17}\)

\(=\dfrac{-1}{7}\left(\dfrac{141}{3}-13\right)\)

\(=-\dfrac{1}{7}\cdot\left(47-13\right)=\dfrac{-34}{7}\)

e: \(\dfrac{1}{7}\cdot\dfrac{5}{12}+\dfrac{1}{7}\cdot\dfrac{5}{3}+\dfrac{1}{7}\cdot\dfrac{2}{12}-\dfrac{-2}{3}\cdot\dfrac{1}{7}\)

\(=\dfrac{1}{7}\left(\dfrac{5}{12}+\dfrac{5}{3}+\dfrac{2}{12}+\dfrac{2}{3}\right)\)

\(=\dfrac{1}{7}\cdot\left(\dfrac{7}{12}+\dfrac{7}{3}\right)\)

\(=\dfrac{1}{12}+\dfrac{1}{3}=\dfrac{5}{12}\)

Do \(OC=OD=CD=R\Rightarrow\Delta OCD\) là tam giác đều

\(\Rightarrow\widehat{COD}=60^0\)

Mà \(\widehat{CAD}=\dfrac{1}{2}\widehat{COD}\) (góc nt và góc ở tâm cùng chắn CD)

\(\Rightarrow\widehat{CAD}=30^0\)

AB là đường kính nên \(\widehat{ADB}\) là góc nt chắn nửa đường tròn \(\Rightarrow\widehat{ADB}=90^0\)

\(\Rightarrow\widehat{ADP}=90^0\Rightarrow\widehat{APB}=180^0-\left(90^0+30^0\right)=60^0\)

Tương tự ta có \(\widehat{ACB}\) là góc nt chắn nửa đường tròn

\(\Rightarrow\widehat{ACB}=90^0\Rightarrow\widehat{BCP}=90^0\)

\(\Rightarrow\widehat{CQD}=360^0-\left(\widehat{APB}+\widehat{ADP}+\widehat{ACB}\right)=360^0-\left(60^0+90^0+90^0\right)=120^0\)

\(\Rightarrow\widehat{AQB}=\widehat{CQD}=120^0\) (2 góc đối đỉnh)

a: x và y tỉ lệ thuận nên \(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\dfrac{x_1+x_2}{y_1+y_2}=\dfrac{6}{-2}=-3\)

=>x=-3y

b: x=-3y

=>\(y=-\dfrac{1}{3}x\)

Thay x=2 vào \(y=-\dfrac{1}{3}x\), ta được:

\(y=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

Thay x=4 vào \(y=-\dfrac{1}{3}x\), ta được:

\(y=-\dfrac{1}{3}\cdot4=-\dfrac{4}{3}\)

a: Thay m=4 vào phương trình, ta được:

\(x^2-4x+4-1=0\)

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b: \(\text{Δ}=\left(-4\right)^2-4\cdot1\left(m-1\right)\)

\(=16-4m+4=-4m+20\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>-4m+20>0

=>-4m>-20

=>\(m< 5\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{\left(-4\right)}{1}=4\\x_1\cdot x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)

\(x_1\left(x_1+2\right)+x_2\left(x_2+2\right)=20\)

=>\(\left(x_1^2+x_2^2\right)+2\left(x_1+x_2\right)=20\)

=>\(\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)=20\)

=>\(4^2-2\cdot\left(m-1\right)+2\cdot4=20\)

=>-2(m-1)+24=20

=>-2(m-1)=-4

=>m-1=2

=>m=3(nhận)

Câu 2:

a: Xét ΔABC có AD là phân giác

nên \(\dfrac{DB}{AB}=\dfrac{DC}{AC}\)

=>\(\dfrac{DB}{15}=\dfrac{DC}{20}\)

=>\(\dfrac{DB}{3}=\dfrac{DC}{4}\)

mà DB+DC=BC=25cm

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{DB}{3}=\dfrac{DC}{4}=\dfrac{DB+DC}{3+4}=\dfrac{25}{7}\)

=>\(DB=\dfrac{25}{7}\cdot3=\dfrac{75}{7}\left(cm\right);DC=\dfrac{25}{7}\cdot4=\dfrac{100}{7}\left(cm\right)\)

Xét ΔCAB có ED//AB

nên \(\dfrac{ED}{AB}=\dfrac{CD}{CB}\)

=>\(\dfrac{ED}{15}=\dfrac{4}{7}\)

=>\(ED=\dfrac{4}{7}\cdot15=\dfrac{60}{7}\left(cm\right)\)

b: Xét ΔABC có \(AB^2+AC^2=BC^2\)

nên ΔABC vuông tại A

=>\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot15\cdot20=10\cdot15=150\left(cm^2\right)\)

Xét ΔABC có AH là đường cao

nên \(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC\)

=>\(AH\cdot25=2\cdot150=300\)

=>\(AH=\dfrac{300}{25}=12\left(cm\right)\)

c: Ta có: \(\dfrac{BD}{BC}=\dfrac{75}{7}:25=\dfrac{3}{7}\)

=>\(S_{ABD}=\dfrac{3}{7}\cdot S_{ABC}=\dfrac{3}{7}\cdot150=\dfrac{450}{7}\left(cm^2\right)\)

=>\(S_{ACD}=S_{ABC}-S_{ABD}=150-\dfrac{450}{7}=\dfrac{600}{7}\left(cm^2\right)\)

Xét ΔCAB có ED//AB

nên \(\dfrac{CE}{CA}=\dfrac{ED}{AB}=\dfrac{CD}{CB}=\dfrac{4}{7}\)

=>\(S_{CED}=\dfrac{4}{7}\cdot S_{CAD}=\dfrac{4}{7}\cdot\dfrac{600}{7}=\dfrac{2400}{49}\left(cm^2\right)\)

Ta có: \(S_{CDE}+S_{AED}=S_{CAD}\)

=>\(S_{AED}=\dfrac{3}{7}\cdot S_{CAD}=\dfrac{3}{7}\cdot\dfrac{600}{7}=\dfrac{1800}{49}\left(cm^2\right)\)

Bài 1:

a: 

b: Để (d1)//(d2) thì \(\left\{{}\begin{matrix}m-2=2\\m\ne1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m=4\\m\ne1\end{matrix}\right.\)

=>m=4

c: Thay x=2 và y=-3 vào (d1), ta được:

\(2\left(m-2\right)+m=-3\)

=>2m-4+m=-3

=>3m=-3+4=1

=>\(m=\dfrac{1}{3}\)

d: Để (d1) cắt (d2) tại một điểm trên trục tung thì

\(\left\{{}\begin{matrix}m-2\ne2\\m=1\end{matrix}\right.\)

=>m=1

Gọi vận tốc riêng của cano là x(km/h) và vân tốc riêng của dòng nước là y (km/h) với x>0,y>0

Vận tốc cano khi xuôi dòng: \(x+y\) (km/h)

Vận tốc cano khi ngược dòng: \(x-y\) (km/h)

Do cano xuôi dòng 84km và ngược dòng 50km hết 5h30 phút =11/2 giờ nên ta có:

\(\dfrac{84}{x+y}+\dfrac{50}{x-y}=\dfrac{11}{2}\)

Do cano xuôi dòng 56km và ngược dòng 60km hết 6h nên:

\(\dfrac{56}{x+y}+\dfrac{60}{x-y}=6\)

Ta được hệ: \(\left\{{}\begin{matrix}\dfrac{84}{x+y}+\dfrac{50}{x-y}=\dfrac{11}{2}\\\dfrac{56}{x+y}+\dfrac{60}{x-y}=6\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}84u+50v=\dfrac{11}{2}\\56u+60v=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{3}{224}\\v=\dfrac{7}{80}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{224}{3}\\x-y=\dfrac{80}{7}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{904}{21}\\y=\dfrac{664}{21}\end{matrix}\right.\)

a: \(\left(\dfrac{1}{32}\right)^7=\left[\left(\dfrac{1}{2}\right)^5\right]^7=\left(\dfrac{1}{2}\right)^{35}\)

\(\left(\dfrac{1}{16}\right)^9=\left[\left(\dfrac{1}{2}\right)^4\right]^9=\left(\dfrac{1}{2}\right)^{36}\)

mà \(\left(\dfrac{1}{2}\right)^{35}>\left(\dfrac{1}{2}\right)^{36}\left(2^{35}< 2^{36}\right)\)

nên \(\left(\dfrac{1}{32}\right)^7>\left(\dfrac{1}{16}\right)^9\)

b: \(\left(\dfrac{1}{243}\right)^9=\left(\dfrac{1}{3}\right)^{5\cdot9}=\left(\dfrac{1}{3}\right)^{45}\)

\(\left(\dfrac{1}{81}\right)^{13}=\left(\dfrac{1}{3}\right)^{4\cdot13}=\left(\dfrac{1}{3}\right)^{81}\)

mà \(\left(\dfrac{1}{3}\right)^{45}>\left(\dfrac{1}{3}\right)^{81}\)

nên \(\left(\dfrac{1}{243}\right)^9>\left(\dfrac{1}{81}\right)^{13}\)

Ta có: 81<83

=>\(\dfrac{1}{81}>\dfrac{1}{83}\)

=>\(\left(\dfrac{1}{81}\right)^{13}>\left(\dfrac{1}{83}\right)^{13}\)

=>\(\left(\dfrac{1}{243}\right)^9>\left(\dfrac{1}{83}\right)^{13}\)

a: \(\dfrac{19}{-23}< \dfrac{-19}{x}< \dfrac{19}{-29}\)

=>\(\dfrac{19}{23}>\dfrac{19}{x}>\dfrac{19}{29}\)

=>\(23< x< 29\)

mà x nguyên dương

nên \(x\in\left\{24;25;26;27;28\right\}\)

b: \(\dfrac{2}{3}< \dfrac{88}{x}< \dfrac{11}{16}\)

=>\(\dfrac{88}{132}< \dfrac{88}{x}< \dfrac{88}{128}\)

=>132>x>128

mà x nguyên dương

nên \(x\in\left\{131;130;129\right\}\)

c: Xét ΔPMN có PQ là đường phân giác

nên \(\dfrac{MQ}{MP}=\dfrac{NQ}{PN}\)

=>\(\dfrac{MQ}{6,2}=\dfrac{QN}{8,7}\)

mà MQ+QN=MN=12,5

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{MQ}{6,2}=\dfrac{QN}{8,7}=\dfrac{MQ+QN}{6,2+8,7}=\dfrac{12.5}{14.9}=\dfrac{125}{149}\)

=>\(\dfrac{x}{8,7}=\dfrac{125}{149}\)

=>\(x=\dfrac{125}{149}\cdot\dfrac{87}{10}=\dfrac{87\cdot25}{2\cdot149}=\dfrac{2175}{298}\)

d: Xét ΔBAC có BD là phân giác

nên \(\dfrac{BA}{BC}=\dfrac{AD}{DC}=\dfrac{4}{5}\)

=>\(\dfrac{BA}{4}=\dfrac{BC}{5}=k\)

=>BA=4k; BC=5k

=>x=4k; y=5k

Ta có: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2-AB^2=AC^2\)

=>\(\left(5k\right)^2-\left(4k\right)^2=9^2\)

=>\(9k^2=81\)

=>\(k^2=9\)

=>k=3

=>\(x=4\cdot3=12;y=5\cdot3=15\)

e: Xét ΔBAC có BD là phân giác

nên \(\dfrac{BA}{BC}=\dfrac{AD}{DC}=\dfrac{2}{3}\)

=>\(\dfrac{BA}{2}=\dfrac{BC}{3}\)

=>\(\dfrac{BA}{4}=\dfrac{BC}{6}\)

Xét ΔCAB có CE là phân giác

nên \(\dfrac{CA}{CB}=\dfrac{AE}{EB}=\dfrac{5}{6}\)

=>\(\dfrac{CA}{5}=\dfrac{CB}{6}\)

=>\(\dfrac{BA}{4}=\dfrac{AC}{5}=\dfrac{BC}{6}\)

mà \(BA+AC+BC=P_{ABC}\cdot2=90\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{BA}{4}=\dfrac{AC}{5}=\dfrac{BC}{6}=\dfrac{AB+AC+BC}{4+5+6}=\dfrac{90}{15}=6\)

=>\(AB=4\cdot6=24\left(cm\right);AC=5\cdot6=30\left(cm\right);BC=6\cdot6=36\left(cm\right)\)