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a: \(\left[600-\left(40:2^3+3\cdot5^3\right)\right]:5\)
\(=\left[600-5-375\right]:5\)
\(=44\)
b: \(16\cdot12^2-\left(4\cdot23^2-59\cdot4\right)\)
\(=16\cdot144-4\cdot\left(23^2-59\right)\)
\(=2304-4\cdot470\)
\(=424\)
c: Ta có: \(2^{100}-\left(1+2+2^2+2^3+...+2^{99}\right)\)
\(=2^{100}-2^{100}+1\)
=1
d: Ta có: \(169\cdot2011^0-17\cdot\left(83-1702:23+1^{2012}\right)+2^7:2^4\)
\(=169-17\cdot\left(83-74+1\right)+2^3\)
\(=177-17\cdot10\)
=7
Bài 3:
a: \(P=x\left(x^2-y\right)+y\left(x-y^2\right)\)
\(=x^3-xy+xy-y^3\)
\(=x^3-y^3\)
Thay \(x=-\frac12;y=-\frac12\) vào P, ta được:
\(P=\left(-\frac12\right)^3-\left(-\frac12\right)^3=\left(-\frac18\right)-\left(-\frac18\right)=-\frac18+\frac18=0\)
b: \(Q=x^2\left(y^3-xy^2\right)+x^2y^2\left(x-y+1\right)\)
\(=x^2y^3-x^3y^2+x^3y^2-x^2y^3+x^2y^2=x^2y^2\)
Thay x=-10; y=-10 vào Q, ta được:
\(Q=\left(-10\right)^2\cdot\left(-10\right)^2=100\cdot100=10000\)
c: \(A=x^3+2xy-2x^3+2y^3+2x^3-y^3\)
\(=\left(x^3-2x^3+2x^3\right)+2xy+\left(2y^3-y^3\right)\)
\(=x^3+2xy+y^3\)
Thay x=2; y=-3 vào A, ta được:
\(A=2^3+2\cdot2\cdot\left(-3\right)+\left(-3\right)^3\)
=8-12-27
=-4-27
=-31
d:
x=1; y=-1
=>\(xy=1\cdot\left(-1\right)=-1\)
\(B=xy+x^2y^2-x^4y^4+x^6y^6-x^8y^8\)
\(=\left(xy\right)+\left(xy\right)^2-\left(xy\right)^4+\left(xy\right)^6-\left(xy\right)^8\)
\(=\left(-1\right)+\left(-1\right)^2-\left(-1\right)^4+\left(-1\right)^6-\left(-1\right)^8\)
=-1+1-1+1-1
=-1
e: x=-1; y=1
=>xy=-1
\(C=xy+x^2y^2+x^3y^3+\cdots+x^{10}y^{10}\)
\(=xy+\left(xy\right)^2+\left(xy\right)^3+\cdots+\left(xy\right)^{10}\)
\(=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+\cdots+\left(-1\right)^{10}\)
=-1+1+(-1)+1+...+(-1)+1
=0
f: \(M=2x^2\left(x^2-5\right)+x\left(-2x^3+4x\right)+x^2\left(x+6\right)\)
\(=2x^4-10x^2-2x^4+4x^2+x^3+6x^2\)
\(=x^3\)
Khi x=-4 thì \(M=\left(-4\right)^3=-64\)
g: \(N=x^3\left(y+1\right)-xy\left(x^2-2x+1\right)-x\left(x^2+2xy-3y\right)\)
\(=x^3y+x^3-x^3y+2x^2y-xy-x^3-2x^2y+3xy\)
=2xy
Thay x=8; y=-5 vào N, ta được:
\(N=2\cdot8\cdot\left(-5\right)=-80\)
Bài 1:
d: \(x^2+2xy-3\cdot\left(-xy\right)\)
\(=x^2+2xy+3xy=x^2+5xy\)
e: \(\frac12x^2y\left(2x^3-\frac25xy^2-1\right)\)
\(=\frac12x^2y\cdot2x^3-\frac12x^2y\cdot\frac25xy^2-\frac12x^2y\)
\(=x^5y-\frac15x^3y^3-\frac12x^2y\)
f: \(\left(-xy^2\right)^2\left(x^2-2x+1\right)\)
\(=x^2y^4\left(x^2-2x+1\right)\)
\(=x^2y^4\cdot x^2-x^2y^4\cdot2x+x^2y^4\)
\(=x^4y^4-2x^3y^4+x^2y^4\)
g: (2xy+3)(x-2y)
\(=2xy\cdot x-2xy\cdot2y+3\cdot x-3\cdot2y\)
\(=2x^2y-4xy^2+3x-6y\)
h: \(\left(xy+2y\right)\left(x^2y-2xy+4\right)\)
\(=x^3y^2-2x^2y^2+4xy+2x^2y^2-4xy^2+8y\)
\(=x^3y^2+4xy-4xy^2+8y\)
i: \(4\left(x^2-\frac12y\right)\left(x^2+\frac12y\right)\)
\(=4\left(x^4-\frac14y^2\right)\)
\(=4\cdot x^4-4\cdot\frac14y^2=4x^4-y^2\)
k: \(2x^2\left(1-3x+2x^2\right)\)
\(=2x^2\cdot1-2x^2\cdot3x+2x^2\cdot2x^2\)
\(=2x^2-6x^3+4x^4\)
l: \(\left(2x^2-3x+4\right)\left(-\frac12x\right)\)
\(=-\frac12x\cdot2x^2+3x\cdot\frac12x-4\cdot\frac12x=-x^3+\frac32x^2-2x\)
m: \(\frac12xy\left(-x^3+2xy-4y^2\right)\)
\(=-\frac12xy\cdot x^3+\frac12xy\cdot2xy-\frac12xy\cdot4y^2\)
\(=-\frac12x^4y+x^2y^2-2xy^3\)
n: \(\frac12x^2y\left(2x^3-\frac25xy^2-1\right)\)
\(=\frac12x^2y\cdot2x^3-\frac12x^2y\cdot\frac25xy^2-\frac12x^2y\)
\(=x^5y-\frac15x^3y^3-\frac12x^2y\)
13: \(\left(-15\right)+8+7\)
\(=\left(-15\right)+\left(8+7\right)\)
=-15+15
=0
14: \(\left(-8\right)+2+6\)
\(=\left(-8\right)+\left(2+6\right)\)
=-8+8
=0
15: \(\left(-1\right)+3-2\)
\(=\left(-1-2\right)+3\)
=-3+3
=0
16: \(25-8-7\)
\(=25-\left(8+7\right)\)
=25-15
=10
17: \(8-2-6\)
\(=8-\left(2+6\right)\)
=8-8
=0
18: \(\left(-12\right)-3+15\)
\(=\left(-12-3\right)+15\)
=-15+15
=0
13: \(\left(\right. - 15 \left.\right) + 8 + 7\)
\(= \left(\right. - 15 \left.\right) + \left(\right. 8 + 7 \left.\right)\)
=-15+15
=0
14: \(\left(\right. - 8 \left.\right) + 2 + 6\)
\(= \left(\right. - 8 \left.\right) + \left(\right. 2 + 6 \left.\right)\)
=-8+8
=0
15: \(\left(\right. - 1 \left.\right) + 3 - 2\)
\(= \left(\right. - 1 - 2 \left.\right) + 3\)
=-3+3
=0
16: \(25 - 8 - 7\)
\(= 25 - \left(\right. 8 + 7 \left.\right)\)
=25-15
=10
17: \(8 - 2 - 6\)
\(= 8 - \left(\right. 2 + 6 \left.\right)\)
=8-8
=0
18: \(\left(\right. - 12 \left.\right) - 3 + 15\)
\(= \left(\right. - 12 - 3 \left.\right) + 15\)
=-15+15
=30
câu d)
\(\dfrac{6}{5}+\left(3+\dfrac{-1}{5}\right)\\ =\dfrac{6}{5}+3+\dfrac{-1}{5}\\ =1+3=4\)
câu e)
\(-\dfrac{3}{5}+\left(-\dfrac{2}{5}+2\right)\\ =\dfrac{-3}{5}+\dfrac{-2}{5}+2\\ =-1+2=1\)
câu f)
\(\dfrac{8}{3}-\left(-\dfrac{4}{3}+3\right)\\ =\dfrac{8}{3}+\dfrac{4}{3}-3\\ =4-3=1\)
Lời giải:
g.
$=\frac{-5}{11}+\frac{-6}{11}+1=(\frac{-5}{11}+\frac{-6}{11})+1$
$=\frac{-11}{11}+1=(-1)+1=0$
h.
$=\frac{-17}{13}+\frac{4}{13}+\frac{25}{101}$
$=\frac{-13}{13}+\frac{25}{101}=(-1)+\frac{25}{101}=-(1-\frac{25}{101})$
$=\frac{-76}{101}$
i.
$=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}$
$=\frac{6}{7}+\frac{1}{8}-\frac{6}{8}=\frac{6}{7}+\frac{-5}{8}$
$=\frac{13}{56}$
CÂU A)
\(\dfrac{5}{3}+\left(7+\dfrac{-5}{3}\right)\\ =\dfrac{5}{3}+7+\dfrac{-5}{3}=7\)
CÂI B)
\(-\dfrac{7}{31}+\left(\dfrac{24}{17}+\dfrac{7}{31}\right)\\ =-\dfrac{7}{31}+\dfrac{24}{17}+\dfrac{7}{31}=\dfrac{24}{17}\)
CÂU C)
\(\dfrac{3}{7}+\left(-\dfrac{1}{5}+\dfrac{-3}{7}\right)\\ =\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
k.
$=\frac{-5}{7}(\frac{2}{11}+\frac{9}{11})+\frac{12}{7}$
$=\frac{-5}{7}.\frac{11}{11}+\frac{12}{7}$
$=\frac{-5}{7}+\frac{12}{7}=\frac{7}{7}=1$
l.
$=(\frac{1}{5}+\frac{4}{5})+(\frac{-2}{9}+\frac{-7}{9})+\frac{16}{17}$
$=\frac{5}{5}+\frac{-9}{9}+\frac{16}{17}$
$=1+(-1)+\frac{16}{17}=\frac{16}{17}$
a.
$=(\frac{-17}{13}+\frac{4}{13})+(\frac{11}{31}+\frac{20}{31})+\frac{2}{135}$
$=\frac{-13}{13}+\frac{31}{31}+\frac{2}{135}=(-1)+1+\frac{2}{135}=\frac{2}{135}$
b.
$=(\frac{3}{17}+\frac{20}{-17})+(\frac{4}{18}+\frac{-2}{9})+(\frac{5}{8}+\frac{21}{56})$
$=(\frac{3}{17}+\frac{-20}{17})+(\frac{2}{9}+\frac{-2}{9})+(\frac{5}{8}+\frac{3}{8})$
$=\frac{-17}{17}+\frac{0}{9}+\frac{8}{8}$
$=(-1)+0+1=0$
c.
$=\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{11}+\frac{5}{12}$
$=(\frac{-5}{12}+\frac{5}{12})+(\frac{6}{11}+\frac{5}{11})+\frac{7}{17}$
$=\frac{0}{12}+\frac{11}{11}+\frac{7}{17}$
$=0+1+\frac{7}{17}=\frac{24}{17}$
d.
$=\frac{9}{16}+\frac{-8}{27}+1+\frac{7}{16}+\frac{-19}{27}$
$=(\frac{9}{16}+\frac{7}{16})+1+(\frac{-8}{27}+\frac{-19}{27})$
$=\frac{16}{16}+1+\frac{-27}{27}$
$=1+1+(-1)=1$
Bài 2 :
a, \(\left(x-1\right)^3=-8\Leftrightarrow\left(x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow x-1=-2\Leftrightarrow x=-1\)
b, \(x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow x=0;x=-1\)
c, \(\left(2x+1\right)^2=25\Leftrightarrow\left(2x+1\right)^2=5^2\)
TH1 : \(2x+1=5\Leftrightarrow x=2\)
TH2 : \(2x+1=-5\Leftrightarrow x=-3\)
d, \(\left(2x-3\right)^2=36\Leftrightarrow\left(2x-3\right)^2=6^2\)
chia 2 trường hợp giống ý c
e, \(5^{x+2}=625\Leftrightarrow5^{x+2}=5^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
f, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left(2-x\right)\left(x+1\right)=0\Leftrightarrow x=1;x=2\)
Bài 3 :
a, Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{7}=\frac{y}{13}=\frac{x+y}{7+13}=\frac{40}{20}=2\)
\(\Leftrightarrow x=14;y=26\)
b, tương tự
c, Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{19}=\frac{y}{21}=\frac{2x-y}{38-21}=\frac{34}{17}=2\)
\(x=38;y=42\)
d, Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)
\(\Leftrightarrow\frac{x^2}{9}=4\Leftrightarrow x^2=36\Leftrightarrow x=6\)
\(\Leftrightarrow\frac{y^2}{16}=4\Leftrightarrow y^2=64\Leftrightarrow y=8\)
d, Theo bài ra ta có : \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\)(*)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}\)(**)
Từ (*) ; (**) suy ra : \(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau
- giải nốt nhé
e, Theo bài ra ta có : \(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)(*)
\(5y=3z\Rightarrow\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)(**)
Từ (*) ; (**) suy ra : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
làm nốt nhé !