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cách 2:
a=\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
a=(6-5-3)-(2/3+5/3-7/3)+(1/2+3/2-5/2)
a=-2-1/2
a=-5/2
Cách 1: = ( 36/6 - 4/6 + 3/6 ) - ( 30/6 + 10/6 - 9/6 ) - ( 18/6 - 14/6 + 15/6 )
= 35/6 - 31/6 - 19/6
= -5/2
Cách 2: = 6 - 2/3 + 1/2 - 5 - 5/3 + 3/2 -3 + 7/3 - 5/2
= ( 6 - 5 - 3 ) + ( -2/3 - 5/3 + 7/3 ) + ( 1/2 + 3/2 - 5/2 )
= -2 + 0 + -1/2
= -5/2
1) 1/3 x 1/2 x 3/7 = 3/42 = 1/14
2) 5/4 x 1/3 +1/7 = 5/12 + 1/7 = 35/84 + 12/84 = 47/84
3) 8 x ( 8/9 - 2/3 ) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 240/240 = 1
5) ( 2/5 + 3/4 ) + 8 = 23/20 + 8 = 23//20 + 160/20 = 183/20
6) 10 x ( 1/2 - 1/5 ) = 10 x 3/10 = 10/1 x 3/10 = 30/10 = 3
Bài 2:
a: \(A=\left(x+1\right)^3+5=20^3+5=8005\)
b: \(B=\left(x-1\right)^3+1=10^3+1=1001\)
5. Ta có: a(a - 1) - (a + 3)(a + 2) = a2 - a - a2 - 2a - 3a - 6
= -6a - 6 = -6(a + 1) \(⋮\)6
<=> -6(a + 1) \(⋮\)6 \(\forall\)a \(\in\)Z
<=> a(a - 1) - (a + 3)(a + 2) \(⋮\) 6 \(\forall\)a \(\in\)Z
6. Thay x = 99 vào biểu thức A, ta có:
A = 995 - 100.994 + 100. 993 - 100.992 + 100 . 99 - 9
A = 995 - (99 + 1).994 + (99 + 1).993 - (99 + 1).992 + (99 + 1).99 - 9
A = 995 - 995 - 994 + 994 + 993 - 993 - 992 + 992 + 99 - 9
A = 99 - 9
A = 90
Vậy ....
Bài 3:
(3x-1)(2x+7)-(x+1)(6x-5)=16.
=> 6x2+21x-2x-7-(6x2-5x+6x-5)=16
=> 6x2+21x-2x-7-6x2+5x-6x+5=16
=> 18x-2=16
=> 18x=16+2
=> 18x=18
=> x=1
Bài 4:
ta có : \(n\left(n+5\right)-\left(n-3\right)\left(n+2\right)=n^2+5n-\left(n^2+2n-3n-6\right)\)
\(=n^2+5n-n^2-2n+3n+6\)
\(=6n+6=6\left(n+1\right)⋮6\)
⇔6(n+1) chia hết cho 6 với mọi n là số nguyên
⇔n(n+5)−(n−3)(n+2) chia hết cho 6 với mọi n là số nguyên
vậy n(n+5)−(n−3)(n+2) chia hết cho 6 với mọi n là số nguyên (đpcm)
Bài 6:
\(A=x^5-100x^4+100x^3-100x^2+100x-9\)
\(\Rightarrow A=x^5-\left(99+1\right)x^4+\left(99+1\right)x^3-\left(99+1\right)x^2+\left(99+1\right)x-9\)
\(\Rightarrow A=x^5-99x^4-x^4+99x^3+x^3-99x^2-x^2+99x+x-9\)
\(\Rightarrow A=\left(x^5-99x^4\right)-\left(x^4-99x^3\right)+\left(x^3-99x^2\right)-\left(x^2-99x\right)+x-9\)
\(\Rightarrow A=x^4\left(x-99\right)-x^3\left(x-99\right)+x^2\left(x-99\right)-x\left(x-99\right)+x-9\)
\(\Rightarrow A=\left(x-99\right)\left(x^4-x^3+x^2-x\right)+x-9\)
Thay 99=x, ta được:
\(A=\left(x-x\right)\left(x^4-x^3+x^2-x\right)+x-9\)
\(\Rightarrow A=x-9\)
Thay x=99 ta được:
\(A=99-9=90\)
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
Bài 3 :
Vì \(\left(x-2\right)^2\ge0\forall x\)
Nên : \(A=\left(x-2\right)^2-4\ge-4\forall x\)
Vậy \(A_{min}=-4\) khi x = 2
B1: lấy máy tính mà tính thôi bạn (nhớ lm theo từng bước)
B2:
a, \(\left|x-\frac{2}{3}\right|-\frac{1}{2}=\frac{5}{6}\)
\(\left|x-\frac{2}{3}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{4}{3}\\x-\frac{2}{3}=\frac{-4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)
b, \(\frac{\left(-2\right)^x}{512}=-32\Rightarrow\left(-2\right)^x=-16384\Rightarrow x\in\varnothing\)
B3:
Vì \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2-4\ge-4\)
Dấu "=" xảy ra khi x = 2
Vậy GTNN của A = -4 khi x = 2
Bài 1:
a, \(\dfrac{-x-2}{3}\) = - \(\dfrac{6}{7}\)
- \(x\) - 2 = - \(\dfrac{18}{7}\)
\(x\) = - 2 + \(\dfrac{18}{7}\)
\(x\) = - \(\dfrac{4}{7}\)
Bài b, \(\dfrac{4}{7-x}\) = \(\dfrac{1}{3}\)
12 = 7 - \(x\)
\(x\) = 7 - 12
\(x\) = -5