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

=> \(x=\frac{1}{3}-\frac{1}{2}=\frac{2}{6}-\frac{3}{6}=\frac{-1}{6}\)

\(x\cdot\frac{1}{3}-\frac{1}{2}=\frac{1}{2}\)

=> \(x\cdot\frac{1}{3}=\frac{1}{2}+\frac{1}{2}=1\)

\(x=1:\frac{1}{3}=3\)

\(x\cdot\frac{1}{2}+x\cdot\frac{1}{3}=\frac{1}{6}\)

=> \(x\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{1}{6}\)

=> \(x\cdot\frac{5}{6}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{5}{6}=\frac{1}{6}\cdot\frac{6}{5}=\frac{1}{5}\)

\(x:\frac{2}{3}+x:\frac{1}{5}=6\)

=> \(x\cdot\frac{3}{2}+x\cdot5=6\)

=> \(x\cdot\left(\frac{3}{2}+5\right)=6\)

=> \(x\cdot\frac{13}{2}=6\)

=> \(x=6:\frac{13}{2}=6\cdot\frac{2}{13}=\frac{12}{13}\)

P/S : Dấu " ." đây là dấu nhân nhé , cấp 2 mới sử dụng

Nhưng mà bạn lớp 4 nên ghi dấu nhân ở trên đề ( có dấu " x " đó)

bạn huỳnh quang sang ơi, sai đề rồi bạn ak

X + 3/2 = 4/3 

nhầm câu đầu nha

1,8 nha bn

nhầm câu hỏi

\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

B=2+22+23+24+...+299+2100=2(1+22+23+24)+...+296(1+22+23+24)=2.31+26.31+...+296.31=31(2+26+...+296)⋮31

\(A=\dfrac{1}{16}c^2-9c+10=\dfrac{1}{16}\left(x-72\right)^2-314\ge-314\)

\(A_{min}=-314\) khi \(c=72\)

\(B=\left(d^2-6de+9e^2\right)+\left(e^2-10e+25\right)+1=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\)

\(B_{min}=1\) khi \(\left\{{}\begin{matrix}d=15\\e=5\end{matrix}\right.\)

\(C=4x^4+12x^2+11\)

Do \(\left\{{}\begin{matrix}x^4\ge0\\x^2\ge0\end{matrix}\right.\) ; \(\forall x\Rightarrow C\ge11\)

\(C_{min}=11\) khi \(x=0\)

a) Ta có: \(\dfrac{1}{16}c^2-9c+10\)

\(=\left(\dfrac{1}{4}c\right)^2-2\cdot\dfrac{1}{4}c\cdot18+324-314\)

\(=\left(\dfrac{1}{4}c-18\right)^2-314\ge-314\forall c\)

Dấu '=' xảy ra khi \(\dfrac{1}{4}c=18\)

hay c=72

Vậy: Giá trị nhỏ nhất của biểu thức \(\dfrac{1}{16}c^2-9c+10\) là -314 khi c=72

b) Ta có: \(d^2+10e^2-6de-10e+26\)

\(=d^2-6de+9e^2+e^2-10e+25+1\)

\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\forall d,e\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}e=5\\d=3e=3\cdot5=15\end{matrix}\right.\)

Vậy: Giá trị nhỏ nhất của biểu thức \(d^2+10e^2-6de-10e+26\) là 1 khi e=5 và d=15

c) Ta có: \(4x^4+12x^2+11\)

\(=4x^4+12x^2+9+2\)

\(=\left(2x^2+3\right)^2+2\ge3^2+2=11\)

Dấu '=' xảy ra khi x=0

Vậy: Giá trị nhỏ nhất của biểu thức \(4x^4+12x^2+11\) là 11 khi x=0

1. likes

2. enjoy

3. welcomed

4. doesn't like

5. works

6. eat - sing

7. don't have 

8. make

9. eat - light 

10. decorate - get 

11. clean

12. does

13. don't like

14. doesn't like

15. to go

16. watch

17. plays - gets

18. starts - ends

19. is - need

20. to eat

yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

<=> 2x + 12 = 3x - 21

<=> 2x - 3x  = -21 - 12

<=>    -x       = -33

<=>      x       = 33

a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)

\(=-5x^3-38x^2-76x-46\)

b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)

\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)

\(=x^3-20x^2+58x+69\)

c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)

\(=-18x^3-46x^2-8x+16\).

a) Ta có: \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)

\(=x^3+2x^2-9x-18-\left(x^3-3x-x^2+3\right)-5x^3-40x^2-80x-x^2+10x-25\)

\(=-4x^3-39x^2-79x-43-x^3+3x+x^2-3\)

\(=-5x^3-38x^2-76x-46\)

b) Ta có: \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2x^2+20x+50-x^2+2x-1\)

\(=2x^3-16x^2+32x-x^3+4x-5x^2+20+x^2+22x+49\)

\(=x^3-20x^2+56x+49\)

c) Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3+48x-36x-2x^3+18x+x^2-9\)

\(=-18x^3+2x^2+40x+16\)

a, Ta có : \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)

\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x-5\right)^2\)

\(=x^3-9x+2x^2-18-\left(x^3-x^2-3x+3\right)-\left(5x^3+40x^2+80x\right)-\left(x^2-10x+25\right)\)

\(=x^3-9x+2x^2-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)

\(=-5x^3-38x^2-76x-46\)

\(b,\) Ta có : \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x-4\right)^2-\left(x+5\right)\left(x^2-4\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)

\(=2x\left(x^2-8x+16\right)-\left(x^3+5x^2-4x-20\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)

\(=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2+20x+50-x^2+2x-1\)

\(=x^3-20x^2+58x+69\)

c, Ta có :\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)

\(=\left(x+5\right)^2-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=\left(x+5\right)^2-\left(16x^3+48x^2+36x\right)-\left(2x^3-x^2-18x+9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+x^2+18x-9\)

\(=-18x^3-46x^2-8x+16\)