a,\(-3x+x-3=0\)
\(\Rightarrow-2x=3\Rightarrow x=-\dfrac{3}{2}\)
b,\(5x\left(x-2\right)-x+2=0\)
\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Chúc bạn học tốt!!!
Tìm \(x\)
a) \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)
Mc: \(x.\left(x-5\right)\)
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - \(x\) - 5 + 5 = 0
\(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0
\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3
Vậy \(x\) = 0 hoặc \(x\) = 3
\(x-5\ne0\Rightarrow x\ne5\)
\(x^2-5\ne0\Rightarrow x\ne5\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {3}
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
Mc: \(x.\left(x+7\right)\)
\(\Leftrightarrow x^2-4x-x-7=-7\)
\(\Leftrightarrow x^2-4x-x=-7+7\)
\(\Leftrightarrow\) \(x^2-5x=0\)
\(\Leftrightarrow x.\left(x-5\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x-5=0\)
\(\Leftrightarrow x=0\) hoặc \(x=5\)
Vậy \(x=0\) hoặc \(x=5\)
\(x+7\ne0\Rightarrow x\ne-7\)
\(x^2+7\ne0\Rightarrow x\ne-7\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {5}
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Mc : \(\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) \(2x^2-4x-4x+8=0\)
\(\Leftrightarrow\) \(2x.\left(x-2\right)-4.\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-4\right).\left(x-2\right)=0\)
\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)
\(\Leftrightarrow x=2\) hoặc \(x=2\)
\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
MC: \(\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)
\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow4.\left(x-1\right)=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)
\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)
\(Mc:\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)
\(\Leftrightarrow0=0\) (Nhận)
Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)
Rút gọn biểu thức sau:
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^2+3\left(x-1\right)\left(x+1\right)\)
b, \(\left(x^4-5x^2+25\right)\left(x^2+5\right)-\left(2+x^2\right)^2+3\left(1+x^2\right)^2\)
Tìm x :
a, \(2\cdot\left(5x+1\right)-7\cdot\left(3x-2\right)=4\cdot\left(2x-1\right)+3\cdot\left(2-x\right)\)
b, \(-4\cdot\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\cdot\left(2x-1\right)+x=5x\cdot\left(1-x\right)\)
Rút gọn các biểu thức sau:
A= \(\left(x+1\right).\left(x^2-x+1\right)+2.\left(x+1\right)-x.\left(x^2+2\right).\)
B= \(\left(5x+1\right).\left(x+7\right)-5x.\left(x-1\right).\)
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) \(\left\{2x-3\left(x-1\right)-5\left[x-4\left(3-2x\right)+10\right]\right\}.\left(-2x\right)\)
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 \(x=1;y=1;z=\left|1\right|\)
1. \(\frac{1}{2}x^2-\left(\frac{1}{2}x-4\right)\frac{1}{2}x=-14\)
2. \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)
3. \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
4. \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
5. \(\left(-2+x^3\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
