\(\dfrac{5^4.18^4}{125.9^5.16}\\ =\dfrac{5^4.\left(2.3^2\right)^4}{5^3.\left(3^2\right)^5.2^4}\\ =\dfrac{5^4.2^4.3^8}{5^3.2^4.3^{10}}\\ =\dfrac{5}{3^2}\\ =\dfrac{5}{9}\)

Câu 1: D

Câu 2: D

Câu 3: A 

Câu 4: A 

Đáp án

Câu 1 : Chọn D

Câu 2 : Chọn D

Câu 3 : Chọn A

Câu 4 : Chọn A

@ChiDung

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{\dfrac{1}{4};1\right\}\end{matrix}\right.\)

\(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\)

\(=\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}-\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt[]{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\dfrac{x-1}{2x+\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}\)

\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(x-2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt[]{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(x-2\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)

Xét tứ giác HMIK có \(\widehat{H}+\widehat{M}+\widehat{I}+\widehat{K}=360^0\)

=>\(3x+4x+2x+x=360\)

=>\(10x=360^0\)

=>\(x=36^0\)

=>\(\widehat{H}=3\cdot36^0=108^0;\widehat{M}=4\cdot36^0=144^0;\widehat{I}=2\cdot36^0=72^0;\widehat{K}=36^0\)

Vì \(\widehat{H}+\widehat{I}=180^0\)

nên HM//IK

=>HMIK là hình thang

\(\dfrac{11}{5}-\left(0,35+x\right)=1\dfrac{1}{2}\\ \dfrac{11}{5}-\left(\dfrac{7}{20}+x\right)=\dfrac{3}{2}\\ \dfrac{11}{5}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{44}{20}-\dfrac{7}{20}-x=\dfrac{3}{2}\\ \dfrac{37}{20}-x=\dfrac{3}{2}\\ x=\dfrac{37}{20}-\dfrac{3}{2}\\ x=\dfrac{7}{20}\)

C

C

Nếu \(a< b\) và \(b< c\) thì \(a< c\) (theo tính chất bắc cầu)

cảm ơn bạn nguyễn tú

1: \(2^3\cdot2^2\cdot2^4=2^{3+2+4}=2^9\)

2: \(2^3\cdot2\cdot2^5=2^{3+1+5}=2^9\)

3: \(10^2\cdot10^3\cdot10^5=10^{2+3+5}=10^{10}\)

4: \(x\cdot x^5=x^{1+5}=x^6\)

5: \(a^3\cdot a^2\cdot a^5=a^{3+2+5}=a^{10}\)

6: \(x^5\cdot x^4\cdot x\cdot x^7\cdot x^6=x^{5+4+1+7+6}=x^{23}\)

7: \(10\cdot10^2=10^{1+2}=10^3\)

8: \(10\cdot100\cdot10^3=10\cdot10^2\cdot10^3=10^6\)

9: \(10\cdot100\cdot10^4\cdot1000=10\cdot10^2\cdot10^4\cdot10^3=10^{10}\)

10: \(5^3:5^2=5^{3-2}=5^1\)

11: \(3^3:3^3=3^{3-3}=3^0\)

12: \(2^7:2^3=2^{7-3}=2^4\)

13: \(4^8:4^4=4^{8-4}=4^4\)

14: \(9^5:9^2=9^{5-2}=9^3\)

15: \(8^9:8^7=8^{9-7}=8^2\)

16: \(a^6:a^3=a^{6-3}=a^3\)

17: \(b^9:b^4=b^{9-4}=b^5\)

khai triển đa thức ta đc:

=x²-4x+4+x²+4x+4+x³+9x²+27x+27+27x³+27x²+9x+1

=28x³+36x²+36x+36

Vậy hệ số của x² sau khi khai triển là 36

a)Xét △HCA và △ACBB

có\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{AHC\left(=90\right)\left(gt\right)}\\\widehat{ACB}chung\end{matrix}\right.\)

⇒△HCA và △ACB (g.g)

b)Có △AHC vuông tại H, HE là đường cao (gt)

⇒EH²=AE.EC ( nhận xét hai △ đồng dạng trong △vuông)