Tìm x
a) 3x(x-1)+\(7x^2\)(x-1)=0
b) \(x^2-2018x-2019=0\)
c) \(\left(x+3\right)^2-x\left(x-2\right)=13\)
Mình đg cần gấp . đảm bảo tick trả đầy đủ :3
Những câu hỏi liên quan
1 . Tìm GTNN của biểu thức :
M = \(x^2-4xy+4y^2-2x+4y+10\)
2 . Tìm x :
a) \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
b) \(\left(3x^3+24\right):\left(x+2\right)+\left(2x^3-54\right):\left(x^2+3x+9\right)=6\)
( Mk đang cần gấp . đảm bảo tick trả đầy đủ )
1 M=\(x^2-4xy+4y^2-2x+4y+10\)
=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)
\(=\left(x-2y\right)\left(x-2y-2\right)+10\)
vì \(\left(x-2y\right)\left(x-2y-2\right)\ge0\)
nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)
\(\Rightarrow\)A\(\ge13\)
dấu "=" xảy ra khi (x-2y)(x-2y-2)=0
\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)
vậy GTNN của M=10 khi x=0; y=0
x=2;y=1
Rút gọn các biểu thức sau :a ) A left(frac{1}{4}x-yright)left(x^2+4xy+16y^2right)+4left(4y^3-frac{1}{16}x^3+1right)b ) B 2xleft(x-4right)^2-left(x+5right)left(x-2right)left(x+2right)+2left(x-5right)^2-left(x-1right)^2c ) C frac{80x^3-125x}{3left(x-3right)-left(x-3right)left(8-4xright)}d ) D frac{left(a-bright)left(c-dright)}{left(b^2-a^2right)left(d^2-c^2right)}Mình đg cần gấp . đảm bảo tick trả đầy đủ
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Rút gọn các biểu thức sau :
a ) A = \(\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
b ) B = \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
c ) C = \(\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
d ) D = \(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
Mình đg cần gấp . đảm bảo tick trả đầy đủ
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !
Tính : \(\frac{1}{\left(x-y\right)\left(y-z\right)}+\frac{1}{\left(y-z\right)\left(z-x\right)}+\frac{1}{\left(z-x\right)\left(x-y\right)}\)
Mk đg cần gấp , đảm bảo tick trả đầy đủ
\(=\frac{z-x}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{x-y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Cho biểu thức A = \(\left(\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right):\frac{x^2-4}{3x^2+6x}\)
a ) Tìm điều kiện của x để A có nghĩa . Rút gọn A
b ) Tìm x nguyên sao cho A nhận giá trị nguyên
( Mình đg cần gấp ạ . Tick trả đầy đủ )
a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)
+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)
\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)
+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)
+) \(x+1\ne0\Leftrightarrow x\ne-1\)
+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)
\(\Leftrightarrow x\ne0;x\ne-2\)
+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)
Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)
a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)
\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)
\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x+1\right)-2x^2-4x+1-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)
b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)
Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x-2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| x | -4 | -1 | 0 | 1 | 3 | 4 | 5 | 8 |
Vậy ............................
b) \(A=\left(\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right):\frac{x^2-4}{3x^2+6x}\)
\(=\left(\frac{\left(x+1\right)^2}{x^2-x+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{x^2-4}{3x^2+6x}\)
\(=\left(\frac{\left(x+1\right)^3}{x^3+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{x^3+1}\right)\)\(.\frac{3x^2+6x}{x^2-4}\)
\(=\left(\frac{x^3+3x^2+3x+1}{x^3+1}-\frac{2x^2+4x-1}{x^3+1}-\frac{x^2-x+1}{x^3+1}\right)\)\(.\frac{3x^2+6x}{x^2-4}\)
\(=\frac{x^3+1}{x^3+1}\)\(.\frac{3x^2+6x}{x^2-4}\)\(=\frac{3x^2+6x}{x^2-4}\)
\(=\frac{3x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x}{x-2}\)
A nguyên\(\Leftrightarrow3x⋮x-2\)
\(\Leftrightarrow3\left(x-2\right)+6⋮x-2\)
Mà \(\left(x-2\right)⋮x-2\Rightarrow6⋮x-2\)
\(\Rightarrow x-2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Lập bảng:
| \(x-2\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(3\) | \(-3\) | \(6\) | \(-6\) |
| \(x\) | \(3\) | \(1\) | \(4\) | \(0\) | \(5\) | \(-1\) | \(8\) | \(-4\) |
Vậy\(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)
Xem thêm câu trả lời
Cho biểu thức A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
a) Rút gọn A
b) Tính A khi x=3 , x=0
c) Tìm x để A=2
Mình đg cần gấp , đảm bảo tick trả đầy đủ
\(ĐKXĐ:x\ne\pm1\)
a) \(A=\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
\(\Leftrightarrow A=\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
\(\Leftrightarrow A=x-1+x+1-3\)
\(\Leftrightarrow A=2x-3\)
b) Thay x = 3 vào A, ta được :
\(A=2.3-3=3\)
Thay x = 0 vào A, ta được :
\(A=2.0-3=-3\)
c) Để A = 2
\(\Leftrightarrow2x-3=2\)
\(\Leftrightarrow2x=5\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy để \(A=2\Leftrightarrow x=\frac{5}{2}\)
Bài 7. Tìm x,biết:
a) x-3x2=0 e) 5x(3x-1)+x(3x-1)-2(3x-1)=0
b) (x+3)2-x(x-2)=13 c) (x-4)2-36=0
d) x2-7x+12=0 g) x2-2018x-2019=0
Bài 8. Tìm x, biết
a) (2x-1)2=(x+5)2 b) x2-x+1/4
c) 4x4-101x2+25=0 d) x3-3x2+9x-91=0
Bài 1.
Tìm x, biết:a, 2x^3-x^2+2x-10b, 2018x-1+2019xleft(1-2018xright)0c,left(x+2right)^3-x^2left(x-6right)-40d,6x^2-left(2x-3right)left(3x+2right)1e,left(x+1right)^3-left(x-1right)left(x^2+x+1right)-20g,7x^2+2x0h,xleft(x+4right)-x^2-6x10i,xleft(x-1right)+2x-20k,left(3x-1right)^2-left(x+5right)^20l,xleft(2x-3right)-2left(3-2xright)0
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Tìm x, biết:
a, \(2x^3-x^2+2x-1=\)0
b, \(2018x-1+2019x\left(1-2018x\right)=0\)
c,\(\left(x+2\right)^3-x^2\left(x-6\right)-4=0\)
d,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\)
e,\(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)
g,\(7x^2+2x=0\)
h,\(x\left(x+4\right)-x^2-6x=10\)
i,\(x\left(x-1\right)+2x-2=0\)
k,\(\left(3x-1\right)^2-\left(x+5\right)^2=0\)
l,\(x\left(2x-3\right)-2\left(3-2x\right)=0\)
Giải các phương trình sau:1, dfrac{x-1}{3}-xdfrac{2x-4}{4}2, left(x-2right)left(2x-1right)x^2-2x3, 3x^2-4x+104, left|2x-4right|05, left|3x+2right|46, left|2x-5right|left|-x+2right|*Giúp mình với mình đg cần gấp ạ T_T
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Giải các phương trình sau:
1, \(\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\)
2, \(\left(x-2\right)\left(2x-1\right)=x^2-2x\)
3, \(3x^2-4x+1=0\)
4, \(\left|2x-4\right|=0\)
5, \(\left|3x+2\right|=4\)
6, \(\left|2x-5\right|=\left|-x+2\right|\)
*Giúp mình với mình đg cần gấp ạ T_T
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
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\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
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tìm x biếta) left(x-2right)^3left(x-3right)left(x^2+3x+9right)+6left(x+1right)^249b)left(x+2right)left(x^2-2x+4right)-xleft(x^2+2right)15c)left(x+3right)^3-xleft(3x+1right)^2+left(2x+1right)left(4x^2-2x+1right)28d)left(x^2-1right)^3-left(x^4+x^2+1right)left(x^2-1right)0Giúp mk vs đc k ạ mk đg cần gấp
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tìm x biết
a) \(\left(x-2\right)^3\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
b)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
c)\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
d)\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
Giúp mk vs đc k ạ mk đg cần gấp
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
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