\(5^{\left(x-2\right).\left(x+3\right)}=1\)
\(-\left(x-y\right)^2=\left(y-3\right)^2\)
\(7^{\left(2x-1\right).\left(3x-1\right)}=1\)
Những câu hỏi liên quan
giải hệ phương trình
1, left{{}begin{matrix}2x^2+3y173x^2-2y6end{matrix}right.
2, left{{}begin{matrix}left|x-1right|+left|y-1right|24left|x-1right|+3left|y-1right|7end{matrix}right.
3, left{{}begin{matrix}3sqrt{x-1}+2sqrt{y}22sqrt{x-1}-sqrt{y}4end{matrix}right.
4 , left{{}begin{matrix}x+y2left|2x-3yright|1end{matrix}right.
5 , left{{}begin{matrix}2x-y1left|x-yright|left|2y-1right|end{matrix}right.
6,left{{}begin{matrix}left(x-3right)left(y+6right)xyleft(x+2right)left(y-2right)xyend{matrix...
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giải hệ phương trình
1, \(\left\{{}\begin{matrix}2x^2+3y=17\\3x^2-2y=6\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-1\right|=2\\4\left|x-1\right|+3\left|y-1\right|=7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=2\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)
4 , \(\left\{{}\begin{matrix}x+y=2\\\left|2x-3y\right|=1\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}2x-y=1\\\left|x-y\right|=\left|2y-1\right|\end{matrix}\right.\)
6,\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+6\right)=xy\\\left(x+2\right)\left(y-2\right)=xy\end{matrix}\right.\)
7 , \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}4x^2-5\left(y+1\right)=\left(2x-3\right)^2\\3\left(7x+2\right)=5\left(2y-1\right)-3x\end{matrix}\right.\)
Thực hiện phép tính
a,\(\left(x-y\right)\left(y^2+y+1\right)+\left(\dfrac{1}{3}x^2y-y\right)\left(2x+y^2\right)\)
b,\(2x^2\left(x-2\right)+3x\left(x^2-x-2\right)-5\left(3-x^2\right)\)
c,\(\left(x-1\right)\left(x-3\right)-\left(4-x\right)\left(2x-1\right)-3x^3+2x-5\)
a: \(=xy^2+xy+x-y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3-2xy-y^3\)
\(=xy^2-xy+x-2y^3-y^2-y+\dfrac{2}{3}x^3y+\dfrac{1}{3}x^2y^3\)
b: \(=2x^3-4x^2+3x^3-3x^2-6x-15+5x^2\)
\(=5x^3-2x^2-6x-15\)
c: \(=x^2-4x+3+\left(x-4\right)\left(2x-1\right)-3x^3+2x-5\)
\(=-3x^3+x^2-2x-2+2x^2-x-8x+4\)
\(=-3x^3+3x^2-11x+2\)
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Tìm x bt1) 3.left(2x-1right).left(3x-1right)-left(2x-3right).left(9x-1right)-3-3-32) 3x-1.left(2x+7right)-x+1.left(6x-5right)x+2-left(x-5right)3) 3xy.left(x+yright)-left(x+yright).left(x^2+y^2+2xyright)+y^3274) 5.x+1.left(1-xright).left(2x^2+3right)0giúp mk vs mai mk đi hk rùi
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Tìm x bt
1) \(3.\left(2x-1\right).\left(3x-1\right)-\left(2x-3\right).\left(9x-1\right)-3=-3\)\(-3\)
2) \(3x-1.\left(2x+7\right)-x+1.\left(6x-5\right)=x+2-\left(x-5\right)\)
3) \(3xy.\left(x+y\right)-\left(x+y\right).\left(x^2+y^2+2xy\right)+y^3=27\)
4) \(5.x+1.\left(1-x\right).\left(2x^2+3\right)=0\)
giúp mk vs mai mk đi hk rùi
Rút gọn các biểu thức :
a, \(\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)\)
b, \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(c,\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
a)\(9x^2+30x+25+9x^2-30x+25-\left(9x^2-2^2\right)\)
=\(9x^2+54\)=\(9\left(x^2+6\right)\)
b)\(2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
=\(8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
=\(x^3-16x^2+25x\)
c)\(\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left(x+y-z-\left(x+y\right)\right)^2\)=\(\left(-z\right)^2\)
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Rút gọn các biểu thức :
a, \(\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)\)
b, \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(c,\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
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Tính
\(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\)
\(2x^2\left(x-2\right)+3x\left(x^2-x-2\right)-5\left(3-x^2\right)\)
\(\left(x-1\right)\left(x-3\right)-\left(4-x\right)\left(2x+1\right)-3x^2+2x-5\)
hệ phương trình
1, left{{}begin{matrix}frac{1}{x+y}+frac{1}{x-y}frac{5}{8}frac{1}{x+y}-frac{1}{x-y}-frac{3}{8}end{matrix}right.
2, left{{}begin{matrix}frac{4}{2x-3y}+frac{5}{3x+y}2frac{3}{3x+y}-frac{5}{2x-3y}21end{matrix}right.
3, left{{}begin{matrix}frac{7}{x-y+2}+frac{5}{x+y-1}frac{9}{2}frac{3}{x-y+2}+frac{2}{x+y-1}4end{matrix}right.
4, left{{}begin{matrix}frac{3}{x}+frac{5}{y}-frac{3}{2}frac{5}{x}-frac{2}{y}frac{8}{3}end{matrix}right.
5 , left{{}begin{matrix}frac{2}{x+y-1}-frac{4}{x-y+1...
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hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
Bài 1: Rút gọn các biểu thức sau:
a) 3x^2 - 2x( 5+ 1,5x) +10
b) 7x ( 4y- x) + 4y( y-7x) - 2( 2y^2 - 3,5x)
c) left{2x-3left(x-1right)-5left[x-4left(3-2xright)+10right]right}.left(-2xright)
Bài 2: Tìm x, biết:
a) 3( 2x -1) - 5( x -3) + 6( 3x -4) 24
b) 2x^2+3left(x^2-1right)5xleft(x+1right)
c) 2xleft(5-3xright)+2xleft(3x-5right)-3left(x-7right)3
d) 3xleft(x+1right)-2xleft(x+2right)-1-x
Bài 3: Tính giá trị của các biểu thức sau:
a)Ax^2left(x+yright)-yleft(x^2+y^2right)+2002...
Đọc tiếp
Bài 1: Rút gọn các biểu thức sau:
a) \(3x^2\) - 2x( 5+ 1,5x) +10
b) 7x ( 4y- x) + 4y( y-7x) - 2( \(2y^2\) - 3,5x)
c) \(\left\{2x-3\left(x-1\right)-5\left[x-4\left(3-2x\right)+10\right]\right\}.\left(-2x\right)\)
Bài 2: Tìm x, biết:
a) 3( 2x -1) - 5( x -3) + 6( 3x -4) = 24
b) \(2x^2+3\left(x^2-1\right)=5x\left(x+1\right)\)
c) \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=3\)
d) \(3x\left(x+1\right)-2x\left(x+2\right)=-1-x\)
Bài 3: Tính giá trị của các biểu thức sau:
a)\(A=x^2\left(x+y\right)-y\left(x^2+y^2\right)+2002\) Với \(x=1;y=-1\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)-\dfrac{11}{20}\) Với \(x=-0,6;y=-0,75\)
Bài 4: Chứng tỏ rằng giá trị của biểu thức sau không phụ thuộc vào giá trị biến:
a) \(2\left(2x+x^2\right)-x^2\left(x+2\right)+\left(x^3-4x+3\right)\)
b) \(z\left(y-x\right)+y\left(z-x\right)+x\left(y+z\right)-2yz+100\)
c) \(2y\left(y^2+y+1\right)-2y^2\left(y+1\right)-2\left(y+10\right)\)
Bài 5: Tính giá trị của biểu thức:
a) \(A=\left(x-3\right)\left(x-7\right)-\left(2x-5\right)\left(x-1\right)\) Với \(x=0;x=1;x=-1\)
b) \(B=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\) Với \(\left|x\right|=2\)
c) \(C=\left(2x+y\right)\left(2z+y\right)+\left(x-y\right)\left(y-z\right)\) Với \(x=1;y=1;z=\left|1\right|\)
Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
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Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
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Bài 2:
c) \(2x(5-3x)+2x(3x-5)-3(x-7)=3\)
\(\Leftrightarrow 2x(5-3x)-2x(5-3x)-3(x-7)=3\)
\(\Leftrightarrow -3(x-7)=3\)
\(\Leftrightarrow x-7=-1\Rightarrow x=6\)
d)
\(3x(x+1)-2x(x+2)=-1-x\)
\(\Leftrightarrow 3x^2+3x-(2x^2+4x)+x+1=0\)
\(\Leftrightarrow x^2+1=0\)
Vô lý vì \(x^2+1\geq 0+1=1>0\) với mọi $x$
Vậy không tồn tại $x$ thỏa mãn.
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Xem thêm câu trả lời
Rút gọn các biểu thức: Aleft(x+3right)left(x^2-3x+9right)-left(54+x^3right)Bleft(2x+yright)left(4x^2-2xy+y^2right)-left(2x-yright)left(4x^2+2xy+y^2right)Cleft(2x+1right)^2+left(1-3xright)^2+2left(2x+1right)left(3x-1right)Dleft(x-2right)left(x^2+2x+4right)-left(x+1right)^3+3left(x-1right)left(x+1right)
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Rút gọn các biểu thức:
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\)
\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\)
\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
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