#2611 help me

131,57 kg thóc

số kg thóc buổi chiều mẹ thu được là:

125 x (1 - 5%) = 118,75 (kg) 

vậy đáp số: 118,75 kg thóc

a, Thay x = 16 ta được 

\(A=\dfrac{16}{4-3}=16\)

b, Với x > 0 ; x khác 9 

\(B=\dfrac{2x-3-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\)

Vậy ta có đpcm 

c, \(A-B< 0\Leftrightarrow\dfrac{x}{\sqrt{x}-3}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-3}< 0\Rightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)

Kết hợp đk vậy 0 < x < 9 

1) Thay `x=16` vào A ta có:

\(A=\dfrac{16}{\sqrt{16}-3}=\dfrac{16}{4-3}=16\)

2) 

\(B=\dfrac{2x-3}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\\ =\dfrac{2x-3}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{2x-3-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\)

3) 

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

Vì: \(\left(\sqrt{x}-1\right)^2\ge0\) nên để \(A-B< 0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)

Kết hợp với đk: 0 < x < 9 

sửa đề \(\dfrac{x-2}{995}-1+\dfrac{x}{997}-1=\dfrac{x-1}{996}-1-\dfrac{x+1}{998}-1\)

\(\Leftrightarrow\dfrac{x-997}{995}+\dfrac{x-997}{997}=\dfrac{x-997}{996}-\dfrac{x-997}{998}\)

\(\Leftrightarrow\left(x-997\right)\left(\dfrac{1}{995}+\dfrac{1}{997}-\dfrac{1}{996}+\dfrac{1}{998}\right)=0\Leftrightarrow x=997\)

7) \(\dfrac{6^{15}\cdot9^{10}}{3^{34}\cdot2^{13}}=\dfrac{2^{15}\cdot3^{15}\cdot\left(3^2\right)^{10}}{3^{34}\cdot2^{13}}=\dfrac{2^{15}\cdot3^{35}}{3^{34}\cdot2^{13}}=2^2\cdot3=12\)

8) \(\dfrac{9^2\cdot2^{11}}{16^2\cdot6^3}=\dfrac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot2^3\cdot3^3}=\dfrac{3^4\cdot2^{11}}{2^{11}\cdot3^3}=3\)

9) \(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)

10) \(\dfrac{27^4\cdot4^3}{9^5\cdot8^2}=\dfrac{\left(3^3\right)^4\cdot\left(2^2\right)^3}{\left(3^2\right)^5\cdot\left(2^3\right)^2}=\dfrac{3^{12}\cdot2^6}{3^{10}\cdot2^6}=3^2=9\)

11) \(\dfrac{3^{29}\cdot4^{16}}{27^9\cdot8^{11}}=\dfrac{3^{29}\cdot\left(2^2\right)^{16}}{\left(3^3\right)^9\cdot\left(2^3\right)^{11}}=\dfrac{3^{29}\cdot2^{32}}{3^{27}\cdot2^{33}}=\dfrac{3^2}{2}=\dfrac{9}{2}\)

7) \(\dfrac{6^{15}.9^{10}}{3^{34}.2^{13}}=\dfrac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}=\dfrac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}\)

\(=\dfrac{2^{15}.3^{35}}{3^{34}.2^{13}}=2^2.3=12\)

8) \(\dfrac{9^2.2^{11}}{16^2.6^3}=\dfrac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.\left(2.3\right)^3}=\dfrac{3^4.2^{11}}{2^8.2^3.3^3}\)

\(=\dfrac{2^{11}.3^4}{2^{11}.3^3}=3\)

9) \(\dfrac{4^5.9^4}{8^3.27^3}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4}{\left(2^3\right)^3.\left(3^3\right)^3}=\dfrac{2^{10}.3^8}{2^9.3^9}=\dfrac{2}{3}\)

10) \(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)

11) \(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}=\dfrac{3^{29}.\left(2^2\right)^{16}}{\left(3^3\right)^9.\left(2^3\right)^{11}}=\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}\)

\(=\dfrac{3^2}{2}=\dfrac{9}{2}\)

12) \(\dfrac{4^{20}.3^{35}}{2^{37}.27^{12}}=\dfrac{\left(2^2\right)^{20}.3^{35}}{2^{37}.\left(3^3\right)^{12}}=\dfrac{2^{40}.3^{35}}{2^{37}.3^{36}}\)

\(=\dfrac{2^3}{3}=\dfrac{8}{3}\)

13) \(\dfrac{6^7.4^2}{9^2.12^5}=\dfrac{\left(2.3\right)^7.\left(2^2\right)^2}{\left(3^2\right)^2.\left(3.2^2\right)^5}=\dfrac{2^7.3^7.2^4}{3^4.3^5.2^{10}}\)

\(=\dfrac{2^{11}.3^7}{2^{10}.3^9}=\dfrac{2}{3^2}=\dfrac{2}{9}\)

14) \(\dfrac{15^2.9^3}{25^3.27^2}=\dfrac{\left(3.5\right)^2.\left(3^2\right)^3}{\left(5^2\right)^3.\left(3^3\right)^2}=\dfrac{3^2.5^2.3^6}{5^6.3^6}\)

\(=\dfrac{3^8.5^2}{5^6.3^6}=\dfrac{3^2}{5^4}=\dfrac{9}{625}\)

15) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.\left(5.2^2\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}=\dfrac{5^4.5^4.2^8}{5^{10}.2^{10}}\)

\(=\dfrac{5^8.2^8}{5^{10}.2^{10}}=\dfrac{1}{5^2.2^2}=\dfrac{1}{100}\)

Gọi số cần tìm có dạng là \(X=\overline{ab}\)

Viết thêm chữ số 2 vào trước số đó thì số mới gấp 5 lần số ban đầu nên ta có: \(\overline{2ab}=5\cdot\overline{ab}\)

=>\(200+X=5X\)

=>4X=200

=>X=50

Vậy: Số cần tìm là 50

Vì nếu viết thêm chữ số 2 vào trước số đó thì được số mới nên số mới hơn số cũ 200 đơn vị.

Hiệu số phần bằng nhau là:

\(5-1=4\) (phần)

Số cần tìm là:

\(200:4\times1=50\)

a: 5 số tự nhiên chẵn liên tiếp mà số nhỏ nhất trong 5 số đó là a+2 là:

a+2;a+4;a+6;a+8;a+10

b: 5 số tự nhiên lẻ liên tiếp mà số lớn nhất trong 5 số đó là a-11 là:

a-11;a-13;a-15;a-17;a-19

\(\left(5^{19}:5^{17}+3\right):7\\ =\left(5^2+3\right):7\\ =28:7\\ =4\)

===========

\(7^9:7^7-3^2+2^3\cdot5^2\\ =7^2-3^2+8\cdot25\\ =49-9+200\\ =40+200\\ =240\)

============

\(1200:2+6^2\cdot2+18\\ =600+36\cdot2+18\\ =600+72+18\\ =600+90\\ =690\)

\(\left(5^{19}:5^{17}+3\right):7=\dfrac{\left(5^2+3\right)}{7}=\dfrac{25+3}{7}=\dfrac{28}{7}=4\)

\(7^9:7^7-3^2+2^3\cdot5^2\)

\(=7^2-3^2+8\cdot25\)

=49-9+200

=240

\(1200:2+6^2\cdot2^1+18\)

=600+36*2+18

=600+90=690

a: \(3x^2+4x+4=0\)

\(\Delta=4^2-4\cdot3\cdot4=16-48=-32< 0\)

=>Phương trình vô nghiệm

b: \(2x^2-x-6=0\)

=>\(2x^2-4x+3x-6=0\)

=>2x(x-2)+3(x-2)=0

=>(x-2)(2x+3)=0

=>\(\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(3x^2+4x+4=0\)

\(\Leftrightarrow3\left(x^2+\dfrac{4}{3}x\right)+4=0\)

\(\Leftrightarrow3\left[x^2+2\cdot x\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2\right]-3\cdot\dfrac{4}{9}+4=0\)

\(\Leftrightarrow3\left(x+\dfrac{2}{3}\right)^2+\dfrac{8}{3}=0\)

\(\Rightarrow x\in\varnothing\) (vì \(3\left(x+\dfrac{2}{3}\right)^2+\dfrac{8}{3}>0\forall x\))

$---$

\(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

#$\mathtt{Toru}$

2008 x 867 + 2009 x 133

= 2008 x 867 + (2008 + 1) x 133

= 2008 x 867 + 2008 x 133 + 133

= 2008 x (867 + 133) + 133

= 2008 x 1000 + 133

= 2008000 + 133

= 2008133 

2008x867+2009x133

=2008x867+2008x133+133

=2008x1000+133

=2008000+133

=2008133