Môn Toán Lớp 7 Mọi ng giúp em vs ????????
Question
Natalia
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Trả lời ( )
Đáp án:
Bài 1:
a.$-x^2y^3z$
b.$-\dfrac1{10}y^4$
c.$-\dfrac15x^4y^5z^2$
Bài 2:
a.$6x^2+6y^2$
b.$4xy-4y^2$
c.$-8x^2+6xy-2y^2$
Bài 3:
a.$ P=-2x^2-9x+4$
b.$-18y^5+15y^4+2$
Bài 4:
$M=4x^2y+\dfrac{14xy^2}{15}-2xy-\dfrac{2}{3}$
Giải thích các bước giải:
Bài 1:
a.Ta có:
$\dfrac15xy^2z\cdot (-5xy)$
$=(\dfrac15\cdot (-5))\cdot (x\cdot x)\cdot (y^2\cdot y)\cdot z$
$=-x^2y^3z$
b.Ta có:
$-\dfrac12y\cdot \dfrac15y^2\cdot y$
$=(-\dfrac12\cdot\dfrac15)\cdot (y\cdot y^2\cdot y)$
$=-\dfrac1{10}y^4$
c.Ta có:
$(\dfrac12x^2y^2)^2\cdot (-\dfrac45yz^2)$
$=(\dfrac12)^2\cdot (x^2)^2\cdot (y^2)^2\cdot (-\dfrac45yz^2)$
$=\dfrac14x^4y^4\cdot (-\dfrac45yz^2)$
$=(-\dfrac45\cdot\dfrac14)\cdot x^4\cdot (y^4\cdot y)\cdot z^2$
$=-\dfrac15x^4y^5z^2$
Bài 2:
a.Ta có:
$A+B+C$
$=(4x^2-5xy+3y^2)+(3x^2+2xy+y^2)+(-x^2+3xy+2y^2)$
$=4x^2-5xy+3y^2+3x^2+2xy+y^2-x^2+3xy+2y^2$
$=4x^2+3x^2-x^2-5xy+2xy+3xy+3y^2+y^2+2y^2$
$=6x^2+6y^2$
b.Ta có:
$B-C-A$
$=(3x^2+2xy+y^2)-(-x^2+3xy+2y^2)-(4x^2-5xy+3y^2)$
$=3x^2+2xy+y^2+x^2-3xy-2y^2-4x^2+5xy-3y^2$
$=3x^2+x^2-4x^2+2xy-3xy+5xy+y^2-2y^2-3y^2$
$=4xy-4y^2$
c.Ta có:
$C-A-B$
$=(-x^2+3xy+2y^2)-(4x^2-5xy+3y^2)-(3x^2+2xy+y^2)$
$=-x^2+3xy+2y^2-4x^2+5xy-3y^2-3x^2-2xy-y^2$
$=-x^2-4x^2-3x^2+3xy+5xy-2xy+2y^2-3y^2-y^2$
$=-8x^2+6xy-2y^2$
Bài 3:
a.Ta có:
$P+(3x^2-4+5x)=x^2-4x$
$\to P=x^2-4x-(3x^2-4+5x)$
$\to P=x^2-4x-3x^2+4-5x$
$\to P=x^2-3x^2-4x-5x+4$
$\to P=-2x^2-9x+4$
b.Ta có:
$Q-14y^4+6y^5-3=-12y^5+y^4-1$
$\to Q=-12y^5+y^4-1+14y^4-6y^5+3$
$\to Q=-12y^5-6y^5+y^4+14y^4-1+3$
$\to Q=-18y^5+15y^4+2$
Bài 4:
Ta có:
$M=x^2y+\dfrac13xy^2+\dfrac35xy^2-2xy+3x^2y-\dfrac23$
$\to M=x^2y+3x^2y+\dfrac{1}{3}xy^2+\dfrac{3}{5}xy^2-2xy-\dfrac{2}{3}$
$\to M=4x^2y+\dfrac{14xy^2}{15}-2xy-\dfrac{2}{3}$
Chúc bn học tốt!!
vote 5*, ctlhn nha