Bạn tham khảo link này :

https://olm.vn/hoi-dap/detail/12023721128.html

~Study well~

#KSJ

Gọi số có 4 chữ số cần tìm là : abcd

Ta có : 41 . abcd = 12abcd

=>       41.abcd = 12000 + abcd

=>       41.abcd - abcd = 120000

=>       40.abcd            = 120000

=>            abcd            = 120000 : 40

=>            abcd            = 3000

Bằng 42

a) Xét ∆AEB và ∆ADC ta có :

EA = AC 

DA = AB 

EAB = DAC( 2 góc đối đỉnh) 

=> ∆AEB = ∆ADC (c.g.c)(dpcm)

=> BE = DC ( 2 cạnh tương ứng) (dpcm)

a)

có \(\widehat{DAC}=90^0+\widehat{BAC}\) ; \(\widehat{BAE}=90^0+\widehat{BAC}\)

\(\Rightarrow\widehat{DAC}=\widehat{BAE}\)

Xét \(\Delta ADC\)và \(\Delta ABE\)

có \(\widehat{DAC}=\widehat{BAE}\)

\(AB=AD\)

\(AC=AE\)

nên \(\Delta ADC=\text{​​}\Delta ABE\left(c-g-c\right)\)

b) 

có\(\Delta ADC=\text{​​}\Delta ABE\)

nên \(CD=BE\)

\(\left(\frac{1}{3}.x\right)\div\frac{2}{3}=1\frac{3}{4}\div\frac{2}{5}\)

\(\Rightarrow\left(\frac{1}{3}.x\right)\div\frac{2}{3}=\frac{7}{4}.\frac{5}{2}\)

\(\Rightarrow\left(\frac{1}{3}.x\right)\div\frac{2}{3}=\frac{35}{8}\)

\(\Rightarrow\frac{1}{3}.x=\frac{35}{8}.\frac{2}{3}\)

\(\Rightarrow\frac{1}{3}.x=\frac{35}{12}\)

\(\Rightarrow x=\frac{35}{12}\div\frac{1}{3}\)

\(\Rightarrow x=\frac{35}{12}.3\)

\(\Rightarrow x=\frac{35}{4}\)

~Study well~

#KSJ

\(100x^2-20x+1\)

\(=\left(10x\right)^2-2.10x.1+1^2\)

\(=\left(10x-1\right)^2\)

\(=\left(0-1\right)^2=1\)

thayx=0, co

100.0^2-20.0+1

=100.0-20.0+1

=0-0+1

=1

ta có 9999= 99 *101. 
do đó 9999^10 = 99 ^10 * 101^10 
còn 99^20 = 99^10 * 99^10 
vì 99^10 < 101^10 nên 99^10 * 99^10 < 99 ^10 * 101^10 . 
vậy 99^20 < 9999^10. 

ta co:9999^10 = 99^10 x 99^10 x 2^10 = 99^200

        suy ra :9999^10=99^200

vay . . .

a.ĐK:\(x\ge0;x\ne1\)

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

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

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

b.\(\frac{3\sqrt{x}}{x-\sqrt{x}+1}< 1\Leftrightarrow3\sqrt{x}< x+\sqrt{x}+1\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)

(Luôn đúng với mọi \(x\ge0;x\ne1\)) . Ta có đpcm

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\sqrt{2}+1\)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

\(x-\frac{3}{4}< \frac{2}{3}\)

\(\Rightarrow x< \frac{2}{3}+\frac{3}{4}\)

\(\Rightarrow x< \frac{8}{12}+\frac{9}{12}\)

\(\Leftrightarrow x< \frac{17}{12}\)

\(x-\frac{3}{4}< \frac{2}{3}\)

\(=>x< \frac{2}{3}+\frac{3}{4}\)

\(=>x< \frac{17}{12}\)