\(\left(x+1\right)\cdot\left(x^2-x+1\right)-\left(x-1\right)\cdot\left(x^2+x+1\right)\)
Những câu hỏi liên quan
Tìm x :
3xcdotleft(x-2right)-2xcdotleft(2x-1right)left(1-xright)cdotleft(1+xright)
left(5x+3right)cdotleft(3x-5right)-left(x-2right)cdotleft(2x+1right)6xcdotleft(3x+1right)-x^2
left(2x-1right)cdotleft(2x+1right)-3cdotleft(x-1right)left(1-4xright)cdotleft(1-xright)
left(2x^2+1right)cdotleft(3x^2-1right)-left(4x^2-3right)cdotleft(x^2+1right)xcdotleft(2x^3+1right)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
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Tìm x :
\(3x\cdot\left(x-2\right)-2x\cdot\left(2x-1\right)=\left(1-x\right)\cdot\left(1+x\right)\)
\(\left(5x+3\right)\cdot\left(3x-5\right)-\left(x-2\right)\cdot\left(2x+1\right)=6x\cdot\left(3x+1\right)-x^2\)
\(\left(2x-1\right)\cdot\left(2x+1\right)-3\cdot\left(x-1\right)=\left(1-4x\right)\cdot\left(1-x\right)\)
\(\left(2x^2+1\right)\cdot\left(3x^2-1\right)-\left(4x^2-3\right)\cdot\left(x^2+1\right)=x\cdot\left(2x^3+1\right)\)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
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1:tìm x
a; 3x+left|x-2right|8
b; 5-left|x-1right|4
2:tìm x
5cdotleft(x-2right)-4cdotleft(1-3xright)left|3-7right|+2cdotleft(1+2xright)
3: tìm x
left(x-2right)cdotleft(2x+1right)-3cdotleft(x+2right)4-5cdotleft(1-xright)
4:tìm x
1dfrac{1}{2}cdotleft(x-2right)-dfrac{x-5}{3}3dfrac{1}{3}cdotleft(1-2xright)-dfrac{5cdotleft(x+1right)}{6}
5: tìm x
left(x-3right)cdotleft(1-xright)+left(x-2right)^2left(1-xright)^2-2cdotleft(1+xright)
6: tìm x
left(2x-1right)^2-3cdotleft(x+2right)^24cdotleft(x...
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1:tìm x
a; \(3x+\left|x-2\right|=8\)
b; \(5-\left|x-1\right|=4\)
2:tìm x
\(5\cdot\left(x-2\right)-4\cdot\left(1-3x\right)=\left|3-7\right|+2\cdot\left(1+2x\right)\)
3: tìm x
\(\left(x-2\right)\cdot\left(2x+1\right)-3\cdot\left(x+2\right)=4-5\cdot\left(1-x\right)\)
4:tìm x
\(1\dfrac{1}{2}\cdot\left(x-2\right)-\dfrac{x-5}{3}=3\dfrac{1}{3}\cdot\left(1-2x\right)-\dfrac{5\cdot\left(x+1\right)}{6}\)
5: tìm x
\(\left(x-3\right)\cdot\left(1-x\right)+\left(x-2\right)^2=\left(1-x\right)^2-2\cdot\left(1+x\right)\)
6: tìm x
\(\left(2x-1\right)^2-3\cdot\left(x+2\right)^2=4\cdot\left(x-2\right)-5\cdot\left(x-1\right)^2\)
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
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2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
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4. 1\(\dfrac{1}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = 3\(\dfrac{1}{3}\).(1 - 2x) - \(\dfrac{5.\left(x+1\right)}{6}\)
<=> \(\dfrac{3}{2}\).(x - 2) - \(\dfrac{x-5}{3}\) = \(\dfrac{10}{3}\).(1 - 2x) - \(\dfrac{5x+5}{6}\)
<=> \(\dfrac{3}{2}x-3-\dfrac{x}{3}+\dfrac{5}{3}=\dfrac{10}{3}-\dfrac{20}{3}x-\dfrac{5x}{6}-\dfrac{5}{6}\)
<=> \(\dfrac{3}{2}x-\dfrac{x}{3}+\dfrac{20}{3}x-\dfrac{5x}{6}=\dfrac{10}{3}-\dfrac{5}{6}-3+\dfrac{5}{3}\)
<=> 7x = \(\dfrac{7}{6}\)
<=> x = \(\dfrac{1}{6}\)
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Xem thêm câu trả lời
Tìm x :
a, \(4x^2-\left(3x+1\right)\cdot\left(2x-1\right)=2\cdot\left(x-3\right)^2\)
b.\(\left(5x-1\right)\cdot\left(x+1\right)-\left(2x-1\right)\cdot\left(2x+1\right)=x\cdot\left(x+1\right)\)
c, \(7x^2-\left(2x-3\right)^2=1+3\cdot\left(x+2\right)^2\)
\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)
\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)
\(\Leftrightarrow-4x^2+13x-17=0\)
\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)
\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)
\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm
\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)
\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)
\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)
\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)
Vậy phương trình vô nghiệm
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2 tháng 10 2020 lúc 16:54
Rút gọn: \(\frac{x^2}{\left(x+y\right)\cdot\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\cdot\left(1+x\right)}-\frac{x^2\cdot y^2}{\left(x+1\right)\cdot\left(1-y\right)}\)
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
Cho mk hỏi con này ra bao nhiu z: \(A=\frac{\left(\sqrt{x}+2\right)\cdot\left(x-1\right)-\left(\sqrt{x}-2\right)\cdot\left(x+1\right)\cdot\left(x+1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\)
Tìm x:
1, \(\left(x-5\right)\cdot\left(x+5\right)-\left(x+3\right)^2=2x-3\)
2,\(\left(2x+3\right)^2+\left(x-1\right)\cdot\left(x+1\right)=5\cdot\left(x+2\right)^2\)
3, \(\left(x-4\right)^3-\left(x-5\right)\cdot\left(x^2+5x+25\right)=\left(x+2\right)\cdot\left(x^2-2x+4\right)-\left(x+4\right)^3\)
1.\(\left(x-5\right).\left(x+5\right)-\left(x+3\right)^2=2x-3\)
\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)
\(\Leftrightarrow-8x-31=0\)
\(\Leftrightarrow x=\dfrac{-31}{8}\)
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\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-\left(x^3-5^3\right)=\left(x^3+2^3\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-x^3+5^3=x^3+2^3-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)
\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)
\(\Leftrightarrow96x=-117\)
\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)
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2. \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2\)
\(\Leftrightarrow4x^2+12x+9+x^2-1=5\left(x^2+4x+4\right)\)
\(\Leftrightarrow4x^2+12x+9+x^2-1=5x^2+20x+20\)
\(\Leftrightarrow4x^2+x^2-5x^2+12x-20x=20-9+1\)
\(\Leftrightarrow-8x=12\)
\(\Leftrightarrow x=\dfrac{-12}{8}=\dfrac{-3}{2}\)
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rút gọn biểu thức sau bằng cách nhanh nhấtA left(a^2+b^2-c^2right)^2-left(a^2-b^2+c^2right)^2-4a^2b^2B left(3x^3+3x+1right)cdotleft(3x^3-3x+1right)-left(3x^3+1right)^2C left(2-6xright)^2+left(2-5xright)^2+2cdotleft(6x-2right)cdotleft(2-5xright)D 5cdotleft(3x-1right)^2+4cdotleft(5x+1right)^2-12cdotleft(5x-2right)left(5x+2right)E left(3x-1right)^2+left(2x+4right)cdotleft(1-3xright)+left(x+2right)^2G left(x-1right)^3+4cdotleft(x+1right)cdotleft(1-xright)+3cdotleft(x-1right)cdotleft(x^2+x+1rig...
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rút gọn biểu thức sau bằng cách nhanh nhất
A = \(\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
B = \(\left(3x^3+3x+1\right)\cdot\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
C = \(\left(2-6x\right)^2+\left(2-5x\right)^2+2\cdot\left(6x-2\right)\cdot\left(2-5x\right)\)
D = \(5\cdot\left(3x-1\right)^2+4\cdot\left(5x+1\right)^2-12\cdot\left(5x-2\right)\left(5x+2\right)\)
E = \(\left(3x-1\right)^2+\left(2x+4\right)\cdot\left(1-3x\right)+\left(x+2\right)^2\)
G = \(\left(x-1\right)^3+4\cdot\left(x+1\right)\cdot\left(1-x\right)+3\cdot\left(x-1\right)\cdot\left(x^2+x+1\right)\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
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Chứng minh giá trị biểu thức không phụ thuộc x :
1, \(\left(2x+1\right)^3-\left(2x-1\right)^3-2\cdot\left(4x+3\right)^2+8\cdot\left(x+3\right)^2\)
2,\(\left(2x+1\right)^2\cdot\left(x-1\right)-2\cdot\left(x-2\right)^3+x\cdot\left(3-2x\right)\cdot\left(3+x\right)-\left(3x-3\right)^2\)
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
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Tìm x
1, \(\left(2x-3\right)\cdot\left(2x+3\right)-4\cdot\left(x+2\right)^2=6\)
2,\(\left(3x+2\right)^2-\left(2x-1\right)\cdot\left(2x+1\right)=5\cdot\left(x-2\right)^2\)
3,\(\left(x+2\right)^2-\left(x+3\right)\cdot\left(x-1\right)=5x\)
1. (2x - 3) . (2x+3) - 4 . (x+ 2)2 = 6
[ ( 2x )2 - 32 ] - 4 . ( x2 + 2.x.2 + 22) = 6
4x2 - 9 - 4 . ( x2 + 4x + 4) = 6
4x2 - 9 - 4x2 - 16x - 16 = 6
-16x -25 = 6
x = \(-\dfrac{31}{16}\)
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