ai giup minh voi mai phai nop roi

câu 1 

xét tích 3 số

=(3a^2.b.c^3).(-2a^3b^5c).(-3a^5.b^2.c^2)

=[3.(-2).(-3)].(a^2.a^3.a^5).(b.b^5.b^2).(c.c^3.c^2)

=18.a^10.b^8.c^5 bé hơn hoặc bằng 0

=>tích 3 số đó không thể cùng âm=>3 số đó ko cùng âm dc

bây giờ mk đi học rùi tí về mk làm típ nhá

a: a>b

=>3a>3b

=>3a+5>3b+5

b: a>b

=>2a>2b

=>2a-3>2b-3>2b-4

Ta có : \(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}\\ Đặt\dfrac{a}{3}=\dfrac{b}{4}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\end{matrix}\right.\\ ThayvàoA,tacó:\)

\(A=\dfrac{2a-5b}{a-3b}-\dfrac{4a+b}{8a-2b}\\ \Leftrightarrow=\dfrac{2\cdot3k-5\cdot4k}{3k-3\cdot4k}-\dfrac{4\cdot3k+4k}{8\cdot3k-2\cdot4k}\\ =\dfrac{6k-20k}{3k-12k}-\dfrac{12k+4k}{24k-8k}\\ =\dfrac{14k}{9k}-\dfrac{16k}{16k}\\ =\dfrac{14}{9}-1\\ =\dfrac{5}{9}\)

giúp mk nha !!!(mk cần gấp)

Ta có : 

\(2a=\frac{a}{\frac{1}{2}};3b=\frac{b}{\frac{1}{3}};5b=\frac{b}{\frac{1}{5}};7c=\frac{c}{\frac{1}{7}}\)

Lại có \(\hept{\begin{cases}\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}\\\frac{b}{\frac{1}{5}}=\frac{c}{\frac{1}{7}}\end{cases}}\Rightarrow\frac{a}{\frac{3}{2}}=b=\frac{c}{\frac{5}{7}}\Leftrightarrow\frac{3a}{\frac{9}{2}}=\frac{7b}{1}=\frac{5c}{\frac{25}{7}}\)

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

\(\frac{3a}{\frac{9}{2}}=\frac{7b}{1}=\frac{5c}{\frac{25}{7}}=\frac{3a-7b+5c}{\frac{9}{2}-1+\frac{25}{7}}=\frac{-30}{\frac{99}{14}}=\frac{-140}{33}\)

\(\Rightarrow\hept{\begin{cases}3a=\frac{-140}{33}\cdot\frac{9}{2}=\frac{-210}{11}\Rightarrow a=\frac{-70}{11}\\7b=\frac{-140}{33}\Rightarrow b=\frac{-20}{33}\\5c=\frac{-140}{33}\cdot\frac{25}{7}=\frac{-500}{33}\Rightarrow c=\frac{-100}{33}\end{cases}}\)

Vậy....

Chắc sai =))

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

a^2 - 2ab - 3b^2 = 0

<=> a^2 - 3ab + ab - 3b^2 = 0

<=> a(a - 3b) + b(a - 3b) = 0

<=> (a - 3b)(a + b) = 0

=> a - 3b = 0 hoặc a + b = 0

=> a = 3b hoặc a = -b

+ Nếu a = 3b

A = (7a+2b)/(2a+b) + (9a-5b)/(2a-b)

A = (7.3b+2b)/(2.3b+b) + (9.3b-5b)/(2.3b-b)

A = 23b/7b + 22b/5b

A = 23/7 + 22/5 = 269/35

+ Nếu a = -b

A = (7a+2b)/(2a+b) + (9a-5b)/(2a-b)

A = (-7b+2b)/(-2b+b) + (-9b-5b)/(-2b-b)

A = -5b/-b + (-14b/-3b)

A = 5 + 14/3 = 29/3

1) Ta có: \(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{c}{5}\)

\(\dfrac{a+2b-c}{2+6-5}=\dfrac{15}{3}=5\)

\(\dfrac{a}{2}=5\) ⇒a=10

\(\dfrac{b}{3}=5\) ⇒b=15

\(\dfrac{c}{5}=5\) ⇒c=25

3) Chu vi hình vuông là

7x4=28(cm)

Nửa chu vi HCN là

28:2=14(cm)

Ta có: \(\dfrac{a}{5}=\dfrac{b}{2}\)

\(\dfrac{a+b}{5+2}=\dfrac{14}{7}=2\)

\(\dfrac{a}{5}=2\) ⇒a=10

\(\dfrac{b}{2}=2\) ⇒b=4

Diện tích HCN là

10x4=40(cm²)