trả lời :entrance exam in grade nine

Gọi số tự nhiên có 4 chữ số cần tìm là abcd

Ta có abcd - ab = 4563

ab x 100 + cd - ab = 4563

=> ab x 99 = 4653 - cd (1)

Vì \(10\le cd\le99\)

=> (1) <=> \(4554\le ab\times99\le4643\)

=> \(46\le ab\le46,8\)

=> ab = 46

Khi đó (1) <=> 46 x 99 = 4653 - cd

=> 4554 = 4653 - cd

=> cd = 99

=> abcd = 4699 

Vậy số cần tìm là 4699

                     ♥ Bài giải ♥

Nếu mua kẹo loại 7500đ/ gói thì mua được số gói là:

       5000 x 18 : 7500 = 12 ( gói )

                                  Đáp số: 12 gói

có số tiền là

5000×18=90000 đồng

nếu mua loại kẹo 7500 thì được số gói là

90000:7500=12gói kẹo

a, \(\left(3x+1\right)\left(3x-1\right)-\left(x-2\right)\left(x^2+2x+4\right)=x\left(6-x^2\right)\)

\(\Leftrightarrow9x^2-3x+3x-1-\left(x^3+2x^2+4x-2x^2-4x-8\right)=6x-x^3\)

\(\Leftrightarrow9x^2-1-\left(x^3-8\right)=6x-x^3\)

\(\Leftrightarrow9x^2-1-x^3+8=6x-x^3\)

\(\Leftrightarrow9x^2-1-x^3+8-6x+x^3=0\)

\(\Leftrightarrow9x^2+7-6x=0\)( vô nghiệm ) 

b, Tương tự 

a, \(\left(3x+1\right)\left(3x-1\right)-\left(x-2\right)\left(x^2+2x+4\right)=x\left(6-x^2\right)\)

\(< =>9x^2-1-\left(x-2\right)\left(x^2+2x+2^2\right)=x\left(6-x^2\right)\)

\(< =>9x^2-1-\left(x^3-2^3\right)=6x-x^3\)

\(< =>9x^2-1-x^3+2^3-6x+x^3=0\)

\(< =>9x^2-6x+7=0\)

\(< =>\left(3x\right)^2-2.3x+1=-6\)

\(< =>\left(3x-1\right)^2=-6\)

Do \(\left(3x-1\right)^2\)luôn luôn lớn hơn hoặc bằng 0

Vậy phương trình trên vô nghiệm

It's milk

My favorite drink is milk

\(8x^2+3x+\left(4x^2+x-2\right)\sqrt{x+4}=4\)

\(\Leftrightarrow\left(4x^2+x-2\right)\sqrt{x+4}=4-3x-8x^2\)

\(\Leftrightarrow\left(4x^2+x-2\right)^2\left(x+4\right)=\left(4-3x-8x^2\right)^2\)

\(\Leftrightarrow\left(4x^2+x-2\right)^2\left(x+4\right)-\left(4-3x-8x^2\right)^2=0\)

ĐKXĐ: \(x\ge-4\)

PT \(\Leftrightarrow\left(\sqrt{x+4}+2\right)\left(2x+1-\sqrt{x+4}\right)\left(2x+\sqrt{x+4}\right)=0\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)

\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)

Khử mẫu : \(12x-12+6x-6=4x+3x-7\)

\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)

\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)

\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)

\(\Leftrightarrow18x-18=7x-7\)

\(\Leftrightarrow18x+7x=18+7\)

\(\Leftrightarrow25x=25\)

\(\Leftrightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)

\(\Leftrightarrow x=1\)

( 198 : 18 - 143 : 13 ) x ( 16996 - 1110 : 30 x 305 )

= ( 11 - 11 ) x ( 16996 - 1110 : 30 x 305 )

= 0 x ( 16996 - 1110 : 30 x 305 )

= 0