Tính nhanh :
a) \(\left(-4\right)\left(+125\right)\left(-25\right)\left(-6\right)\left(-8\right)\)
b) \(\left(-98\right)\left(1-246\right)-246.98\)
Tính nhanh :
a) \(\left(-4\right).\left(+3\right).\left(-125\right).\left(+25\right).\left(-8\right)\)
b) \(\left(-65\right).\left(1-301\right)-301.67\)
a (-4) . (+3) . (-125) . (+25) . (-8)
=[(-4) . (+25)] . [(-125) . (-8)] . (+3)
=(-100) + (1000) . (+3)
= -300 000
b (-67) . (1 - 301) - 301 . 67
=(-67) . 1 + 67 . 301 - 67 . 301
= - 67
\(A=\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right):\left(-4\dfrac{1}{6}+3\dfrac{1}{7}\right)+7\dfrac{1}{2}\)
\(B=4\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}:\dfrac{125}{64}\right):\left(-\dfrac{27}{8}\right)\)
giải hộ mk nhanh nhanh nhoa ☺
Giúp mik với
Tính nhanh:
a. A=\(\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\left(n\in N\right)\)
b. B=\(\left(10000-1^2\right)\left(10000-2^2\right)\left(10000-3^2\right)..\left(10000-1000^2\right)\)
c. C=\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
d. D=\(1999^{\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)}\)
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
C1. Kết quả của phép tính \(\left(\dfrac{4}{25}\right)^2.\left(\dfrac{2}{5}\right)^6:\left(\dfrac{-8}{125}\right)^8\) là :
A. \(\dfrac{-2}{5}\) B. \(\dfrac{2}{5}\) C. \(\dfrac{4}{25}\) D. -1
C2. Kết quả của phép tính \(\dfrac{15}{19}.\dfrac{2}{3}-\dfrac{7}{19}.\dfrac{2}{3}+\dfrac{8}{3}.\dfrac{17}{19}\) là :
A. \(\dfrac{17}{19}\) B. \(\dfrac{19}{3}\) C. \(\dfrac{8}{3}\) D. -1
C3. Cho | 3x + 2 | = | 5x - 6 | . Tích các giá trị của x thỏa mãn đẳng thức đã cho là :
A.2 B.4 C.\(\dfrac{1}{2}\) D. 8
C4. Nhà nước trích tiền ủng hộ miền trung khắc phục hậu quả cơn bão số 9 thành ba đợt lần lượt tỉ lệ với 7;8;9 . Biết rằng tổng số tiền đợt hai và đợt ba nhiều hơn đợt một là 80 tỉ . Số tiền ủng hộ đợt hai là :
A. 56 tỉ B.64 tỉ C.72 tỉ D.80 tỉ .
Viết dưới dạng tự luận giúp mk nha mn , thankk .
Viết hết tất cả dưới dạng tự luận luôn à bạn?
C1. Mik ko bik vì bấm máy tính thì nó ra kết quả quá lớn.
C2. \(\dfrac{15}{19}.\dfrac{2}{3}-\dfrac{7}{19}.\dfrac{2}{3}+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{2}{3}.\left(\dfrac{15}{19}-\dfrac{7}{19}\right)+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{2}{3}.\dfrac{8}{19}+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{16}{57}+\dfrac{136}{57}\\ =\dfrac{152}{57}\\ =\dfrac{8}{3}\left(C\right)\)
C3. \(x=4\left(B\right)\)
C4. Gọi tổng số tiền ba đợt ủng hộ lần lượt là a, b, c (a, b, c ϵ N*).
Vì số tiền ủng hộ ba đợt lần lượt tỉ lệ với 7; 8; 9
Nên \(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{9}\)
Vì tổng số tiền ủng hộ đợt hai và đợt ba nhiều hơn số tiền ủng hộ đợt một 80 tỉ
Nên \(\left(b+c\right)-a=80\)
Theo tính chất của dãy tỉ số bằng nhau
Ta có:
\(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{9}=\dfrac{\left(b+c\right)-a}{\left(8+9\right)-7}=\dfrac{80}{10}=8\)
Do đó:
\(\dfrac{a}{7}=8=>a=8.7=>a=56\\ \dfrac{b}{8}=8=>b=8.8=>b=64\\ \dfrac{c}{9}=8=>c=8.9=>c=72\)
Vậy số tiền ủng hộ ba đợt lần lượt là: 56; 64; 72.
Số tiền ủng hộ đợt hai là 64 tỉ (B).
Tính:
a/\(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(\dfrac{-1}{6}\right):\left(-3\right)\)
b/\(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1
\(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-\dfrac{2}{4}-\dfrac{1}{4}\right).\left(-\dfrac{1}{5}\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{3}\right)\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{1}{5}\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{3}{20}+\dfrac{1}{48}-\dfrac{1}{18}=\dfrac{108}{720}+\dfrac{15}{720}-\dfrac{40}{720}=\dfrac{83}{720}\)
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
\(I\)Tính nhanh
\(a.127^2+146.127+73^2\)
\(b.9^8.2^8\left(18^4-1\right)\left(18^4+1\right)\)
\(c.100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(d.\frac{780^2-220^2}{125^2+150.125+75^2}\)
\(II.\)Rút gọn các biểu thức
\(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
Tính nhanh : A= \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{3^3}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(0\)
Bài 1: Tính:
A=\(\left(-2\right).\left(-3\right)-5.\left|-5\right|+125.\left(\dfrac{-1}{5}\right)^2\)
B=\(\left(-3\right).\left|-7\right|-\left(-4\right).\left|5\right|+\dfrac{1}{3}.\left|-9\right|\)
C=\(\left(-2\right)^3.\left|-3\right|-\dfrac{1}{5}.\left|-25\right|-4.\left|-7\right|+\left(-2\right)^2\)
D=\(\left(-6\right).\left|-3\right|+2.\left|-9\right|-7\left|\left(-2\right)^3\right|+8.\left|-7\right|\)
E=\(\left|-3^2\right|.\left|4\right|-\left|7\right|.8-\left|6\right|.\left|-8\right|-\left|12\right|.\left(\dfrac{1}{2}\right)^2\)
Bài 2: Tìm x:
a)\(12-2\left|3x+2\right|=10\)
b)\(2.\left|5-4x\right|+17=\left(-2\right)^3.\left(-4\right)\)
c)\(\left|3x-5\right|+\left(-3\right)^2.2=12.\left|3x+5\right|+117\)
d)\(4.\left|3-2x\right|+\left(-5\right).\left|4-3x\right|-5=-6\)
e)\(\left|2x-7\right|-2^3.\left|2x-7\right|+15=-5.\left|2x-7\right|+3\)
f)\(\left|x+2\right|+\left|x^2-4\right|=0\)
g)\(\left|3x-9\right|+\left|x^2-9\right|=0\)
h)\(\left|2x-1\right|+\left|x^2-\dfrac{1}{4}\right|=0\)
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
1. B = (-3).|-7| - (-4).|5| + \(\dfrac{1}{3}.\left|-9\right|\)
= -21 + 20 + 3
= 2
@Shine Anna