a: Để hệ có nghiệm duy nhất thì \(\dfrac{3}{2}\ne\dfrac{a}{1}\)

=>\(a\ne1,5\)

b: Để hệ vô nghiệm thì \(\dfrac{3}{2}=\dfrac{a}{1}\ne\dfrac{5}{b}\)

=>\(\left\{{}\begin{matrix}a=\dfrac{3}{2}\\b\ne\dfrac{10}{3}\end{matrix}\right.\)

c: Để hệ có vô số nghiệm thì \(\dfrac{3}{2}=\dfrac{a}{1}=\dfrac{5}{b}\)

=>\(\left\{{}\begin{matrix}a=1\cdot\dfrac{3}{2}=\dfrac{3}{2}\\b=5\cdot\dfrac{2}{3}=\dfrac{10}{3}\end{matrix}\right.\)

Bài 7:

a: \(\dfrac{11}{13}=\dfrac{110}{130};\dfrac{12}{13}=\dfrac{120}{130}\)

=>4 phân số nằm giữa 11/13 và 12/13 là \(\dfrac{111}{130};\dfrac{112}{130};\dfrac{113}{130};\dfrac{116}{130}\)

b: \(\dfrac{15}{17}=\dfrac{150}{170};\dfrac{15}{16}=\dfrac{150}{160}\)

=>5 phân số nằm giữa 15/17 và 15/16 là \(\dfrac{150}{169};\dfrac{150}{168};\dfrac{150}{167};\dfrac{150}{165};\dfrac{150}{153}\)

Bài 11:

\(a)x\times12,8=6,4\\ x=6,4:12,8\\ x=\dfrac{1}{2}\\ b)17,3:x=69,2\\ x=17,3:69,2\\ x=\dfrac{1}{4}\\ c)16,48\times x=4,12\\ x=4,12:16,48\\ x=\dfrac{1}{4}\\ d)x:12,8=1,6\\ x=12,8\times1,6\\ x=\dfrac{512}{25}\)

Bài 11:

a: \(x\times12,8=6,4\)

=>\(x=\dfrac{6.4}{12.8}=\dfrac{1}{2}=0,5\)

b: \(17,3:x=69,2\)

=>\(x=\dfrac{17.3}{69.2}=0,25\)

c: \(16,48\times x=4,12\)

=>\(x=\dfrac{4.12}{16.48}=\dfrac{1}{4}=0,25\)

d: \(x:12,8=1,6\)

=>\(x=12,8\times1,6=20,48\)

Bài 9:

a: \(4,86\times0,25\times40=4,86\times10=48,6\)

b: \(0,125\times6,94\times80=6,94\times\left(80\times0,125\right)\)

=6,94x10

=69,4

c: \(96,28\times3,527+3,527\times3,72\)

=3,527x(96,28+3,72)

=3,527x100=352,7

d: \(72,9\times99+72+0,9\)

=72,9x99+72,9

=72,9x(99+1)

=72,9x100=7290

e: \(0,8\times96+1,6\times2\)

\(=0,8\times96+0,8\times4=0,8\times\left(96+4\right)\)

=0,8x100=80

\(D=x^2+y^2-x+6y+10\\ =\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ =\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(y^2+2\cdot y\cdot3+3^2\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.=>D=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)

Dấu "=" xảy ra \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

______________________________

\(F=2xy-2x^2-y^2+10x-27\\ =-\left(x^2-2xy+y^2\right)-\left(x^2-10x+25\right)-2\\ =-\left(x-y\right)^2-\left(x-5\right)^2-2\)

Ta có: \(\left\{{}\begin{matrix}\left(x-y\right)^2\le0\forall x,y\\-\left(x-5\right)^2\le0\forall x\end{matrix}\right.=>F=-\left(x-y\right)^2-\left(x-5\right)^2-2\le-2\forall x,y\)

Dấu "=" xảy ra: \(\left\{{}\begin{matrix}x-y=0\\x-5=0\end{matrix}\right.\Leftrightarrow x=y=5\)

\(A=-x^2+x-1\)

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

\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

=>\(x=\dfrac{1}{2}\)

\(B=6x-x^2-10\)

\(=-\left(x^2-6x+10\right)\)

\(=-\left(x^2-6x+9+1\right)\)

\(=-\left(x-3\right)^2-1< =-1\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

\(C=-x^2+5x+3\)

\(=-\left(x^2-5x-3\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}< =\dfrac{37}{4}\forall x\)

Dấu '=' xảy ra khi x-5/2=0

=>x=5/2

\(D=x^2-x+y^2+6y+10\)

\(=x^2-x+\dfrac{1}{4}+y^2+6y+9+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

\(F=2xy-2x^2-y^2+10x-27\)

\(=-\left(2x^2+y^2-2xy-10x+27\right)\)

\(=-\left(x^2-2xy+y^2+x^2-10x+25+2\right)\)

\(=-\left(x-y\right)^2-\left(x-5\right)^2-2< =-2\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=x=5\end{matrix}\right.\)

Bài 1:

\(a)\left(\dfrac{1}{3}:x-1\right)=\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{21}\\ \dfrac{1}{3}:x-1=\dfrac{7}{21}+\dfrac{3}{21}-\dfrac{1}{21}=\dfrac{9}{21}\\ \dfrac{1}{3}:x-1=\dfrac{3}{7}\\ \dfrac{1}{3}:x=\dfrac{3}{7}+1=\dfrac{10}{7}\\ x=\dfrac{1}{3}:\dfrac{10}{7}\\ x=\dfrac{7}{30}\\ b)\dfrac{1}{5}\cdot x-\dfrac{2}{13}=\dfrac{1}{2\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{5}{8\cdot13}\\ \dfrac{1}{5}\cdot x-\dfrac{2}{13}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}\\ \dfrac{1}{5}\cdot x-\dfrac{2}{13}=\dfrac{1}{2}-\dfrac{1}{13}=\dfrac{11}{26}\\\dfrac{1}{5}\cdot x=\dfrac{11}{26}+\dfrac{2}{13}=\dfrac{15}{26}\\ x=\dfrac{15}{26}:\dfrac{1}{5}=\dfrac{75}{26}\\c)\dfrac{13}{6} :\left(\dfrac{1}{2}+x\right)=\dfrac{1}{3}+\dfrac{3}{7}+\dfrac{1}{7\cdot2}+\dfrac{5}{2\cdot13}+\dfrac{3}{13\cdot4}\\ \dfrac{13}{6}:\left(\dfrac{1}{2}+x\right)=\dfrac{1}{3}+\left(\dfrac{3}{7}+\dfrac{1}{7\cdot2}\right)+\left(\dfrac{5}{2\cdot13}+\dfrac{3}{13\cdot4}\right)\\ \dfrac{13}{6}:\left(\dfrac{1}{2}+x\right)=\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{4}\\ \dfrac{13}{6}:\left(\dfrac{1}{2}+x\right)=\dfrac{13}{12}\\ \dfrac{1}{2}+x=\dfrac{13}{6}:\dfrac{13}{12}=2\\ x=2-\dfrac{1}{2}\\ x=\dfrac{3}{2}\)

bằng 9/10 số hs còn lại nhé

`#3107.101107`

`c,`

`(x^3)/2 + 4`

`= (x^3)/2 + 8/2`

`= (x^3 + 8)/2`

`= 1/2*(x^3 + 8)`

`= 1/2*(x + 2)(x^2 - 2x + 4)`

`d,`

`27y^3 + 27y^2 + 9y + 1`

`= (3y)^3 + 3 * (3y)^2 * 1 + 3 * 3y * 1^2 + 1^3`

`= (3y + 1)^3`

____

HĐT:

`A^3 + B^3 = (A + B)(A^2 - AB + B^2)`

`(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3.`

Gọi mẫu số của phân số cần tìm là x

Theo đề, ta có: \(-\dfrac{11}{13}< \dfrac{9}{x}< \dfrac{-11}{15}\)

=>\(\dfrac{11}{13}>\dfrac{-9}{x}>\dfrac{11}{15}\)

=>\(\dfrac{99}{117}>\dfrac{-99}{11x}>\dfrac{99}{135}\)

=>\(\dfrac{99}{117}>\dfrac{99}{-11x}>\dfrac{99}{135}\)

=>\(-11x\in\left\{118;119;...;134\right\}\)

=>\(x\in\left\{-\dfrac{118}{11};-\dfrac{119}{11};...;\dfrac{134}{-11}\right\}\)

mà x nguyên

nên \(x\in\left\{-11;-12\right\}\)

Vậy: Hai phân số cần tìm là \(\dfrac{9}{-11};\dfrac{9}{-12}\)

=>

\(\dfrac{a}{d}+\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{c}{d}\\ =>\dfrac{a}{b}\cdot\dfrac{c}{d}-\dfrac{c}{d}=\dfrac{a}{b}\\ =>\dfrac{c}{d}\cdot\left(\dfrac{a}{b}-1\right)=\dfrac{a}{b}\\ =>\dfrac{c}{d}\cdot\dfrac{a-b}{b}=\dfrac{a}{b}\\ =>\dfrac{c}{d}=\dfrac{a}{b}:\dfrac{a-b}{b}\\ =>\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{b}{a-b}\\ =>\dfrac{c}{d}=\dfrac{a}{a-b}\)

Vậy: ... 

a) Để A là phân số thì \(n-2\ne0\Rightarrow n\ne2\)

b) \(A=-3=>\dfrac{2n-7}{n-2}=-3\)

\(=>2n-7=-3\left(n-2\right)\\ =>2n-7=-3n+6\\ =>2n+3n=6+7\\ =>5n=13\\ =>n=\dfrac{13}{5}\left(ktm\right)\) 

c) \(A=\dfrac{2n-7}{n-2}=\dfrac{2n-4-3}{n-2}=\dfrac{2\left(n-2\right)-3}{n-2}=2-\dfrac{3}{n-2}\)

Để A nguyên thì: 3 ⋮ n - 2

=> n - 2 ∈ Ư(3) ={1; -1; 3; -3}

=> n ∈ {3; 1; 5; -1} 

d) Để A lớn nhất thì \(\dfrac{3}{n-2}\) nhỏ nhất 

=> \(\dfrac{3}{n-2}=-1\)

=>  3 = -(n - 2)

=> 3 = -n + 2

=> n = -1  

e) Để A nhỏ nhất thì \(\dfrac{3}{n-2}\) lớn nhất

=> \(\dfrac{3}{n-2}=1\)

=> 3 = n - 2

=> n = 3 + 2 

=> n = 5

f) Để A là phân số tối giản => ƯCLN(2n - 7; n - 2) = 1

=> ƯCLN(3; n - 2) = 1

=> n - 2 không chia hết cho 3

=> n - 2 ≠ 3k 

=> n ≠ 3k + 2

g) Gọi d là ước nguyên tố của 2n - 7 và n - 2 ta có: 

2n - 7 ⋮ d và n - 2 ⋮ d 

=> 2n - 7 ⋮ d và 2(n - 2) ⋮ d

=> (2n - 4)  - (2n - 7) ⋮ d

=> 3 ⋮ d 

=> d ∈ {1; -1; 3; -3}

Mà d là STN => d = 3 

Với d = 3 => 2n - 7 ⋮ 3 => 2(2n - 7) ⋮ 3 => 4n - 7 ⋮ 3 

=> 3n + n - 7 ⋮ 3 

=> n - 7 ⋮ 3 

=> n - 7 = 3k 

=> n = 3k + 7 

bạn cho mình hỏi sao câu d và câu e lại là -1 và 1 thế ạ?