Luôn luôn nhe

always:luôn luôn

chúc bsvv

mn ơi  giúp em

Bài 3:

\(a,=3x\left(y-4x+6y^2\right)\\ b,=5xy\left(x^2-6x+9\right)=5xy\left(x-3\right)^2\\ d,=\left(x+y\right)\left(x-12\right)\\ f,=2x\left(x-y\right)\left(5x-4y\right)\\ g,=\left(x-2\right)\left(x-2+3x\right)=\left(x-2\right)\left(4x-2\right)=2\left(x-2\right)\left(2x-1\right)\\ h,=x^2\left(1-5x\right)+3xy\left(5x-1\right)=x\left(1-5x\right)\left(x-3y\right)\\ i,=x\left(x-2\right)+4\left(x-2\right)=\left(x+4\right)\left(x-2\right)\\ j,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ k,=4x^2-12x+3x-9=\left(x-3\right)\left(4x+3\right)\\ l,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ m,=x^2-\left(2y-6\right)^2=\left(x-2y+6\right)\left(x+2y-6\right)\\ n,=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-25\\ =\left(x^2+5x\right)\left(x^2+5x+10\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)

Chu vi hình chữ nhật :

( dài + rộng ) \(\times2\)

Đó là: ( Chiều dài + chiều rộng ) x 2

16A 17B 18B đây nha chị

A

B

B

\(h'\left(x\right)=f'\left(x\right)-g'\left(x\right)=0\Rightarrow x=\left\{a;b;c\right\}\)

Ta thấy \(h'\left(x\right)>0\) trên \(\left(b;c\right)\) và \(h'\left(x\right)< 0\) trên \(\left(a;b\right)\)

\(\Rightarrow x=b\) là điểm cực tiểu trên \(\left[a;c\right]\) hay \(\min\limits_{\left[a;c\right]}h\left(x\right)=h\left(b\right)\)

 \(=>Qthu1=0,2.340000=68000J\)

\(=>Qthu2=2100.0,2.20=8400J\)

\(=>Qtoa=2.4200.25=210000J\)

\(=>Qthu1+Qthu2< Qtoa\)=>đá nóng chảy hoàn toàn

\(=>0,2.2100.20+0,2.340000+0,2.4200.tcb=2.4200\left(25-tcb\right)\)

\(=>tcb=14,5^oC\)

Cho em hỏi ngu tí ạ vậy tcb ở nhưng phép tính trên vứt đi đâu ạ 

a: \(A=4-\sin^245^0+2\cdot cos^260^0-3\cdot\cot^345^0\)

\(=4-\left(\frac{\sqrt2}{2}\right)^2+2\cdot\left(\frac12\right)^2-3\cdot1^3\)

\(=4-\frac12+2\cdot\frac14-3=4-3=1\)

b: \(B=tan45^0\cdot cos30^0\cdot\cot30^0\)

\(=1\cdot\frac{\sqrt3}{2}\cdot\sqrt3=\frac32\)

c: \(C=cos^215^0+cos^225^0+\cdots+cos^275^0\)

\(=\left(cos^215^0+cos^275^0\right)+\left(cos^225^0+cos^265^0\right)+\left(cos^235^0+cos^255^0\right)+cos^245^0\)

\(=1+1+1+\left(\frac{\sqrt2}{2}\right)^2=3+\frac12=\frac72\)

d: \(D=\sin^210^0+\sin^220^0+\cdots+\sin^280^0\)

\(=\left(\sin^210^0+\sin^280^0\right)+\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0+\sin^260^0\right)+\left(\sin^240^0+\sin^250^0\right)\)

=1+1+1+1

=4

Ban can cau nao nhi?

làm bài nào??