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I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
a/\(-5-\left(-3\right)+\left(-4\right)-\left(2-4+3\right)\)
= -5+3 -4- 1
= -7
b/\(2-\left(+4\right)-\left(-3\right)+\left(-5+7\right)\)
= 2 -4 +3 +2
= 3
c/ \(-5-\left(-5\right)-\left(-4+8\right)\)
= -5 +5 -4
= -4
\(\left(-x+4\right)\left(x^2+4x+16\right)\)
= -x3- 4x2-16x+4x2+16x+64
= -x3+64 =(-x+4).( -x2-4x+42)
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
a)\(\left(\frac{1}{5}\right)^{3x-1}=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{1}{5}\right)^{3x-1}=\left(\frac{1}{5}\right)^2\)
<=> 3x-1=2
<=> 3x=3
<=> x=1
c) \(\left(\frac{2}{3}\right)^{1-x}=\left(\frac{2}{3}\right)^3\)
<=> 1-x=3
<=>x=-2
d) (0,7)3x+1=(0,7)3
<=> 3x+1=3
<=> 3x=2
<=> x=2/3
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
BÀi 2:
Cả 4 câu áp dụng tính chất này: \(\sqrt{a^2}=a\)
a)\(\sqrt{\frac{3^2}{7^2}}=\frac{3}{7}\)
b)\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{92^2}}=\frac{3+39}{7+92}=\frac{42}{99}=\frac{14}{33}\)
c)\(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\frac{3-39}{7-91}=\frac{-36}{-84}=\frac{3}{7}\)
d)\(\sqrt{\frac{39^2}{91^2}}=\frac{39}{91}=\frac{3}{7}\)
b)Vì BCNN(3;5) = 15
\(\Rightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{2.5}=\frac{y}{3.5}=\frac{x}{10}=\frac{y}{15};\frac{y}{5}=\frac{z}{7}\Leftrightarrow\frac{y}{5.3}=\frac{z}{7.3}=\frac{y}{15}=\frac{z}{21}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.10=20\\y=2.15=30\\z=2.21=42\end{matrix}\right.\)
Vậy...
c)Vì BCNN(2;3;5) = 30
\(\Rightarrow2x=3y=5z\Leftrightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}=\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
WTFFFFFF>>>
d)dễ... áp dụng tính chất DTBN là ra 1/2 rồi tính
e)Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(x=\frac{y}{2}=\frac{z}{4}=\frac{4x}{4}=\frac{3y}{6}=\frac{2x}{8}=\frac{4x-3y+2x}{4-6+8}=\frac{36}{6}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.1=6\\y=6.2=12\\z=6.4=24\end{matrix}\right.\)
Vậy...
1.
a) -5 - (-5) - (-4 - 8)
= -5 + 5 + 12
= 0 + 12
= 12.
Mình chỉ làm bài 1 thôi nhé.
Chúc bạn học tốt!
Câu 2 :
\(a,\left(-x+4\right)\left(x^2+4x+14\right)\)
=> \(-x^3-4x^2-141x+4x^2+16x+564\)
=> \(-x^3-\left(4x^2-4x^2\right)-\left(141x-16x\right)+564\)
=> \(-x^3-125x+564\)
\(b,3x^2\left(-5x+4y\right)+5xy\left(-3+2\right)\)
=> \(-15x^3+12x^2y+5xy.\left(-1\right)\)
=> \(-15x^3+12x^2y-5xy\)
\(c,4xy\left(3x^2-5\right)-3y\left(4x^3-5yx\right)\)
=> \(12x^3y-20xy-3y\left(4x^3-5xy\right)\)
=> \(12x^3y-20xy-12x+15xy^2\)
=> \(\left(12x^3y-12x^3y\right)-20xy+15xy^2\)
=> \(-20xy+15xy^2\)
#~ Hết~#
a,-200 x10 t10z3
b,\(\frac{-5}{4}\)x11 y5 z4
c,\(\frac{2}{15}\)x6 y6 z9
d,\(\frac{1}{7}\)x10 y6 z7
e,-4z6 y10 z6
Bài 1 :
a) \(-3+\left(-4\right)-\left(-3\right)+\left(2+7-10\right)=-3-4+3+2+7-10=-5\)
b) \(3-\left(-3+2-7\right)+\left(-4\right)=3+3-2+7-4=7\)
c) \(7+\left(-2-3+7\right)-\left(-2\right)=7-2-3+7+2=17\)
d) \(-\left(-3\right)-\left(-2+3-8\right)+\left(-6\right)=3+2-3+8-6=4\)
Bài 2 :
a) \(x^2-2x-\left(3x-2x\right)=x^2-2x-3x+2x=x^2-3x\)
b) \(-\left(x^2+3x^2\right)-\left(-5x^2+3x\right)=-x^2-3x^2+5x^2-3x=x^2-3x\)
c) \(\left(x-y\right)-\left(x+3y+1\right)=x-y-x-3y-1=-4y-1\)
Bài 1:
a, -3+ (-4) - (-3) + (2 + 7 - 10)
= -3 - 4 + 3 + 2 + 7 - 10
= 5 - 10
= -5.
b, 3 - (-3 + 2 - 7) + (-4)
= 3 + 3 - 2 + 7 - 4
= 11 - 4
= 7
c, 7 + (-2 - 3 + 7) - (-2)
= 7 - 2 - 3 + 7 + 2
= 9 + 2
= 11.
d, - (-3) - (-2 + 3 - 8) + (-6)
= 3 + 2 - 3 + 8 - 6
= 10 - 6
= 4.
Mình chỉ làm bài 1 thôi nhé.
Chúc bạn học tốt!