Giup minh voi cau nay voi:
a,1x2x3x4/x5x6x7x8
b,23+23x2+23x3+23x4
c,2x1/2x3x1/3x4x1/4x5x1/5
d,248x2005-205x148
1,tính:
a,1/2x2/3x3/4= b,7/8x8/9x9/10=
c,5/14x7/15x28/7= d,2x1/2x3x1/3x4x1/4x5x1/5=
2,tính:
a,7/20-(5/8-3/5)= b,5/6+(5/9-1/4)=
c,9/10-(2/5+3/10)+7/20=
3,tìm x:
a,1/2+x=5/6 b,5/6-x=1/3 c,x-1/=3/10 d,1/3x x=1/6
Bài 1: Tính
a, 1/2 x 2/3 x 3/4 = 1/4
b, 7/8 x 8/9 x 9/10 = 7/10
c, 5/14 x 7/15 x 28/7 = 2/3
d, 2 x 1/2 x 3 x 1/3 x 4 x 1/4 x 5 x 1/5 = 1
Bài 2: Tính
a, 7/20 - ( 5/8 - 3/5 ) = 7/20 - 1/40 = 14/40 - 1/40 = 13/40
b, 5/6 + ( 5/9 - 1/4 ) = 5/6 + 11/36 = 30/36 + 11/36 = 41/36
c, 9/10 - ( 2/5 + 3/10 ) + 7/20 = 9/10 - 7/10 + 7/20 = 2/10 + 7/20 = 4/20 + 7/20 = 11/20.
Bài 3: Tìm x:
a, 1/2 + x = 5/6
x = 5/6 - 1/2
x = 1/3
b, 5/6 - x = 1/3
x = 5/6 - 1/3
x = 1/2
c, x - 1/mấy vậy bạn
d, 1/3 x x = 1/6
x = 1/6 : 1/3
x = 1/2
Tính bằng cách thuận tiện nhất :
25x12x4x3
23+23x2+23x3+23x4 248x2005-2005x148
3 tính bằng cách thuận tiện
25x12x30x4
23+23x2+23x4+23x4
248x2005 - 2005x148
1/
25x12x30x4
= (25x4)x(12x30)
=100x3600
=360 000
2/
23+23x2+23x3+23x4
=23x1+23x2+23x3+23x4
=23x(1+2+3+4)
=23x10
=230
3/
248x2005 - 2005x148
=(248-148)x2005
=100x2005
=200500
1/
25x12x30x4
= (25x4)x(12x30)
=100x3600
=360 000
2/
23+23x2+23x3+23x4
=23x1+23x2+23x3+23x4
=23x(1+2+3+4)
=23x10
=230
3/
248x2005 - 2005x148
=(248-148)x2005
=100x2005
=200500
1/
25x12x30x4
= (25x4)x(12x30)
=100x3600
=360 000
2/
23+23x2+23x3+23x4
=23x1+23x2+23x3+23x4
=23x(1+2+3+4)
=23x10
=230
3/
248x2005 - 2005x148
=(248-148)x2005
=100x2005
=200500
TBC cua 2 so la 30.Biet St1 gap 4 lan St2. Tim 2 so do
giup minh tra loi cau hoi nay voi ban minh dat thu thach cau hoi minh hong bit cau nay giup minh tra loi cau nay voi bi thua la danh sml lun.
Tổng 2 số: 30*2=60
Số thứ 1: 60/(4+1)*4=48
Số thứ 2: 60-48=12
Vậy St1=48, St2=12
Tổng 2 số là :
30 x 2 = 60
Ta có sơ đồ :
St1 :|___|___|___|___| tổng : 60
St2 :|___|
Tổng số phần bằng nhau là :
4 + 1 = 5 (phần)
St1 là :
60 : 5 x 4 = 48
St2 là :
60 - 48 = 12
Đ/S : st1 : 48
st2 : 12
Cho ti le ban do 1:100.000,san truong co dien h. 36m2. Vay dt that cua san truong bao nhieu?
giup minh cau nay voi
1) \(B=\dfrac{\sqrt{x}-5}{\sqrt{x}}\)
Thay \(x=\dfrac{4}{25}\) vào B, ta được:
\(B=\dfrac{\sqrt{\dfrac{4}{25}}-5}{\sqrt{\dfrac{4}{25}}}\)
\(=\dfrac{\dfrac{2}{5}-5}{\dfrac{2}{5}}\)
\(=\dfrac{-\dfrac{23}{5}}{\dfrac{2}{5}}\)
\(=-\dfrac{23}{2}\)
2) ĐKXĐ: \(x\ne9;x\ge0\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x+9\sqrt{x}}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
3) \(P=A.B\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}.\dfrac{\sqrt{x}-5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}\)
\(=1-\dfrac{2}{\sqrt{x}+3}\)
Để P nhỏ nhất thì \(\dfrac{8}{\sqrt{x}+3}\) lớn nhất
Ta có:
\(\dfrac{8}{\sqrt{x}+3}\ge\dfrac{8}{3}\)
\(\Rightarrow P\) nhỏ nhất là \(1-\dfrac{8}{3}=-\dfrac{5}{3}\) khi \(x=0\)
giup minh cau nay voi
giup minh cau nay voi
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{7}{2}\end{matrix}\right.\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\dfrac{37}{4}\)
\(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\dfrac{153}{8}\)
\(C=x_1^4+x_2^4=\left(x_1^2+x_2^2\right)^2-2\left(x_1x_2\right)^2=\dfrac{977}{16}\)
\(D=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\dfrac{\sqrt{65}}{2}\)
\(E=\left(2x_1+x_2\right)\left(2x_2+x_1\right)=2\left(x_1^2+x_2^2\right)+5x_1x_2=1\)
giup minh cau nay voi
`3x^2+10x+3=0`
Ptr có: `\Delta'=5^2-3.3=16 > 0`
`=>` Ptr có `2` nghiệm pb
`=>` Áp dụng Viét có: `{(x_1+x_2=[-b]/a=-10/3),(x_1 .x_2=c/a=1):}`
~~~~~~~~~~~~~
`A=x_1 ^2+x_2 ^2`
`A=(x_1+x_2)^2-2x_1 .x_2`
`A=(-10/3)^2-2.1=82/9`
_______________________________________________________
`B=x_1 ^3+x_2 ^3`
`B=(x_1+x_2)(x_1 ^2-x_1 .x_2+x_2 ^2)`
`B=(x_1+x_2)[(x_1+x_2)^2 -3x_1 .x_2]`
`B=(-10/3).[(-10/3)^2-3.1]=-730/27`
_______________________________________________________
`C=x_1 ^4+x_2 ^4`
`C=(x_1 ^2+x_2 ^2)^2 -2x_1 ^2 .x_2 ^2`
`C=[(x_1+x_2)^2-2x_1 .x_2]^2-2(x_1 .x_2)^2`
`C=[(-10/3)^2-2.1]^2-2. 1^2=6562/81`
_______________________________________________________
`D=|x_1-x_2|`
`D=\sqrt{(x_1-x_2)^2}`
`D=\sqrt{(x_1+x_2)^2-4x_1.x_2}`
`D=\sqrt{(-10/3)^2-4.1}=8/3`
_______________________________________________________
`E=(2x_1+x_2)(2x_2+x_1)`
`E=4x_1 .x_2+2x_1 ^2+2x_2 ^2+x_1 .x_2`
`E=5x_1 . x_2+2(x_1+x_2)^2-4x_1 .x_2`
`E=x_1 .x_2+2(x_1+x_2)^2`
`E=1+2(-10/3)^2=209/9`
giup minh cau nay voi
`a)` Thay `x=-3` vào ptr có:
`(-3)^2-6.(-3)+2m+1=0`
`<=>9+18+2m+1=0`
`<=>m=-14`
`b)` Ptr có: `\Delta'=(-3)^2-(2m+1)=9-2m-1=8-2m`
Ptr có `2` nghiệm phân biệt `<=>\Delta' > 0`
`<=>8-2m > 0<=>m < 4`
`c)` Ptr có nghiệm kép `<=>\Delta' =0`
`<=>8-2m=0<=>m=4`