71+65.=x-260

331=x+140:x+260

331=x: [140+260]

331=x:400

x=331 nhân 400

x=132 400

   hok tốt

x = 2

* Cách giải lên lớp 6 học *

1. What your favourite food ?

 => My favourite food is pizza.

2. Where are you going this summer ?

 => I'm going to Sam Son Beach.

3. Where are you yesterday morning ?

  => I was at home.

What your favorite food ?

  ->My favorite food is ...

Where are you going this summer ?

 -> I going to ... this summer

where are you yester day morning ?

 -> Cau tra loi có nhiều cách khác nhau

Những dấu ... là bạn tự điền

Bài làm:

Ta có: \(\frac{3+a^2}{b+c}+\frac{3+b^2}{c+a}+\frac{3+c^2}{a+b}\)

\(=\frac{3}{b+c}+\frac{a^2}{b+c}+\frac{3}{c+a}+\frac{b^2}{c+a}+\frac{3}{a+b}+\frac{c^2}{a+b}\)

\(=3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)\)

Áp dụng bất đẳng thức Cauchy Schwars ta được:

\(VT\ge3.\frac{\left(1+1+1\right)^2}{a+b+b+c+c+a}+\frac{\left(a+b+c\right)^2}{b+c+c+a+a+b}\)

\(=3.\frac{9}{2\left(a+b+c\right)}+\frac{3^2}{2\left(a+b+c\right)}\)

\(=3.\frac{9}{2.3}+\frac{9}{2.3}=\frac{9}{2}+\frac{9}{6}=6\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

Sô phần bể Hai vòi cùng chảy trong 1 giờ được là:

1:6=1/6(bể)

Nếu mỗi vòi đều chảy riêng trong 2giờ thỉ cả hai vòi chảy được số phần bể là:

1616×2=2323 (bể)

Vòi thứ hai chảymột mình trong 1giờ là:

2525-2323=115115

Vòi thứ hai chảy một mình đên khi đầy bể là:

1:115115=1×15=15(giờ)

Đáp số:15 giờ

công thức: S1+ ĐT khuyết thiếu+be+Vp2+[by+S2]+O

1.Traffic rules must be obeyed strictly to reduce traffic accidents (by us).

2.Everything should be done well by me before the exam.

3. Sugar shouldn't be used too much in your diets.

4. Posters could be drawn for decorating streets by me.

Bg

c) 9 < 3^x : 3 < 81

=> 3² < 3^{x - 1} < 3⁴ 

=> x - 1 = {2; 3; 4}

=> x = {3; 4; 5}

d) 5^x . 5^{x + 1} . 5 ^{x + 2} < 2¹⁸ . 5¹⁸ : 2¹⁸ 

=> 5^{x + x + 1 + x + 2} < 2¹⁸ : 2¹⁸ . 5¹⁸ 

=> 5^{3x + 3} < 1.5¹⁸ 

=> 5^{3.(x + 1)} < 5¹⁸ 

=> 3.(x + 1) < 18

=> x + 1 < 18 : 3

=> x + 1 < 6

=> x < 6 - 1

=> x < 5

c. \(9\le3^x:3\le81\)

\(\Rightarrow3^2\le3^{x-1}\le3^4\)

\(\Rightarrow3^{x-1}\in\left\{3^2;3^3;3^4\right\}\)

\(\Rightarrow x-1\in\left\{2;3;4\right\}\)

\(\Rightarrow x\in\left\{3;4;5\right\}\)

d. Thêm đk : x thuộc N

 \(5^x.5^{x+1}.5^{x+2}\le2^{18}.5^{18}:2^{18}\)

\(\Rightarrow5^{x+x+1+x+2}\le5^{18}\)

\(\Rightarrow x+x+x+1+2\le18\)

\(\Rightarrow3x+3\le18\)

\(\Rightarrow3\left(x+1\right)\le18\)

\(\Rightarrow x+1\le6\)

\(\Rightarrow x\le5\)

\(\Rightarrow x\in\left\{1;2;3;4;5\right\}\)

Bài 1

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

Khi đó:

\(x^3+x\left(1-a\right)+a=bx^2+cx+2\)

Do đó \(1-a=c;a=2;b=0\Rightarrow a=2;b=0;c=-1\)

Bài 2:

\(A=\left(n^2+2n-5\right)\left(n+2\right)-2n^3+n+10\)

\(=n^3+2n^2+2n^2+4n-5n-10-2n^3+n+10\)

\(=-n^3+4n^2\)

\(=n^2\left(4-n\right)\)

Lập luận với n chẵn thì cái trên luôn chia hết cho 8

1. ( x² - x + a )( x + 1 ) = x³ + bx² + cx + 2

<=> x³ + x² - x² - x + ax + a = x³ + bx² + cx + 2

<=> x³ + 0x² + ( a - 1 )x + a = x³ + bx² + cx + 2

<=> \(\hept{\begin{cases}b=0\\a-1=c\\a=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=0\\c=1\end{cases}}\)

2. n chẵn => n có dạng 2k ( \(k\inℕ^∗\))

Thế vào ta được :

A = [ ( 2k )² + 2.2k - 5 )( 2k + 2 ) - 2(2k)³ + 2k + 10 

A = ( 4k² + 4k - 5 )( 2k + 2 ) - 16k³ + 2k + 10

A = 8k³ + 16k² - 2k - 10 - 16k³ + 2k + 10

A = -8k³ + 16k² = -8k²(k-2) \(⋮\)8

=> A chia hết cho 8 với mọi n chẵn ( đpcm )

Tìm x

\(x^2=36\)

\(x^2=6^2=\left(-6\right)^2\)

\(\Rightarrow x=\pm6\)

Vậy \(x=\pm6\).

\(3x^3=81\)

\(x^3=81\div3\)

\(x^3=27\)

\(x^3=3^3\)

\(\Rightarrow x=3\)

Vậy \(x=3\).

\(\left(4x\right)^2=64\)

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

\(\Rightarrow\orbr{\begin{cases}4x=8\\4x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy \(x=\pm2\).

\(\left(x-2\right)^2=121\)

\(\left(x-2\right)^2=11^2=\left(-11\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=11\\x-2=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=-9\end{cases}}\)

Vậy \(x\in\left\{13;-9\right\}\).

\(a,x^2=36\)

\(\Rightarrow x^2=6^2\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

\(b,3x^3=81\)

\(\Rightarrow x^3=81:3\)

\(\Rightarrow x^3=27\)

\(\Rightarrow x^3=3^3\)

\(\Rightarrow x=3\)

\(c,\left(4x\right)^2=64\)

\(\Rightarrow\left(4x\right)^2=8^2\)

\(\Rightarrow\orbr{\begin{cases}4x=8\\4x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

\(d,\left(x-2\right)^2=121\)

\(\Rightarrow\left(x-2\right)^2=11^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=11\\x-2=-11\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=-9\end{cases}}\)

Học tốt

\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right):\left(1+8+15+...+281+288\right)=\frac{1}{3}\)

\(\Rightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right):\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)

\(\Rightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right):6069=\frac{1}{3}\)

\(\Rightarrow\frac{1488x-279+279-476,5x}{10}=2023\)

\(\Rightarrow1488x-476,5x=20230\)

\(\Rightarrow1011,5x=20230\)

\(\Rightarrow x=20\)

             Bài làm :

Ta có :

\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right)\div\left(1+8+15+...+281+288\right)=\frac{1}{3}\)

\(\Leftrightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right)\div\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)

\(\Leftrightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right)\div6069=\frac{1}{3}\)

\(\Leftrightarrow\frac{1488x-279+279-476,5x}{10}=2023\)

\(\Leftrightarrow1488x-476,5x=20230\)

\(\Leftrightarrow1011,5x=20230\)

\(\Leftrightarrow x=20\)

Vậy x=20