\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=\left(a-3b+a+3b\right)\left(a-3b-a-3b\right)-\left(ab-2a-b+2\right)\)

\(=2a.\left(-6b\right)-ab+2a+b-2\)

\(=-12ab-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

Thay a=1/2,b=-3 vào N ta được:

\(N=-13\cdot\frac{1}{2}.\left(-3\right)+2\cdot\frac{1}{2}-3-2=\frac{39}{2}+\frac{2}{2}-5=\frac{41}{2}-5=\frac{31}{2}\)

rút gọn và tính giúp mình

\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(\Rightarrow N=\left(a-3b\right)^2-\left[\left(a-3b\right)+6b\right]^2-\left(ab-2a-b+2\right)\)

\(\Rightarrow N=\left(a-3b\right)^2-\left(a-3b\right)^2-2\left(a-3b\right)6b-36b^2-ab+2a-b+2\)

\(\Rightarrow N=\left(-2a+6b\right)6b-36b^2-ab+2a-b+2\)

\(\Rightarrow N=-12a+36b^2-36b^2-ab+2a-b+2\)

\(\Rightarrow N=-10a-ab-b+2\)

Thay a = 1/2 ; b = -3 vào ta có

\(\Rightarrow N=-10.\frac{1}{2}-\frac{1}{2}.-3+3+2\)

\(\Leftrightarrow N=-5+\frac{3}{2}+5=\frac{3}{2}\)

Vt lại đề nhé (khó nhìn)

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=x\Rightarrow a=bx;c=dx\)

Lần lượt thay vào các vế, ta được :

\(\dfrac{5a+3b}{5a-3b}=\dfrac{5.b.x+3b}{5.b.x+3b}=\dfrac{b\left(5x+3\right)}{b\left(5x+3\right)}=\dfrac{5x+3}{5x+3}\left(1\right)\)

\(\dfrac{5c-3d}{5c-3d}=\dfrac{5.d.x-3d}{5.d.x-3d}=\dfrac{d\left(5x-3\right)}{d\left(5x-3\right)}=\dfrac{5x-3}{5x-3}\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\)

\(\Rightarrow\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\left(đpcm\right)\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

vãi cả loz sao lại sai ?

Chắc đề ghi nhầm ngoặc sau (2 mẫu kia thực chất giống nhau, lẽ ra phải là \(\dfrac{1}{\sqrt{a+3b}}+\dfrac{1}{\sqrt{3a+b}}\)

\(VT=\sqrt{\dfrac{a}{a+3b}}+\sqrt{\dfrac{a}{3a+b}}+\sqrt{\dfrac{b}{a+3b}}+\sqrt{\dfrac{b}{3a+b}}\)

\(=\sqrt{\dfrac{a}{a+b}.\dfrac{a+b}{a+3b}}+\sqrt{\dfrac{1}{2}.\dfrac{2a}{3a+b}}+\sqrt{\dfrac{1}{2}.\dfrac{2b}{a+3b}}+\sqrt{\dfrac{b}{a+b}.\dfrac{a+b}{3a+b}}\)

\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a+b}{a+3b}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{2a}{3a+b}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{2b}{a+3b}\right)+\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{a+b}{3a+b}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{a+b}{a+b}+\dfrac{a+3b}{a+3b}+\dfrac{3a+b}{3a+b}\right)=2\)

Dấu "=" xảy ra khi \(a=b\)

a)  de sai 

b) do a/b =c/d  =>a/c =b/d =k (1) => k^2 = a.c /bd

tu (1) =>k^2 =a^2/ c^2 =b^2/ d^2 =a^2+b^2 /c^2+d^2 

=>a^2 +b^2 /c^2 +d^2 = a.c /bd

Sử dụng mối liên hệ giữa thứ tự và phép nhân, phép cộng, chúng ta thu được

a) -3a + 4 < -3b + 4;        b) 2 - 3a < 2 - 3b.

\(=\left(b-a\right)\left[\left(a+3b\right)+\left(a-b\right)^2\cdot\left(a+b\right)\right]\)