Rút gọn biểu thức:
\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)
RÚT GỌN BIỂU THỨC
\(2x.\left(2x-1\right)-3x.\left(x+3\right).\left(x-3\right)-4x.\left(x+1\right)^2\)
cho hình thang giác vuông ABCD có
;góc A= D (=90) độ gọi M là trung điểm của bc
CMR: BAM=CDM
làm giúp mình ik mình lm cho
cho hình thang giác vuông ABCD có
;góc A= D (=90) độ gọi M là trung điểm của bc
CMR: BAM=CDM
lm giúp mình ikminhf lm cho
=4x2-2x-3x3-27x-4x3-8x2-4x
=-7x3-33x-4x2
=-x.(7x2+4x+33)
rút gọn biểu thức( không tính cái gạch ở cuối nha)\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)
\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4
rút gọn rồi tính giá trị biểu thức tại x=1; y=2
A= \(\dfrac{6x^3-4x^2y+2x^2}{x\left(3x+y\right)\left(3x-y\right)}\)
\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)
Cho biểu thức C=\(\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\) Với \(x\ne\pm1\)
a, Rút gọn C
b, Tính giá trị của C khi x=\(-1\dfrac{1}{2}\)
c, Tìm x biết C=\(\dfrac{1}{2}\)
d, Chứng minh 2C<1
a) Ta có: \(C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)
\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{\left(1+x\right)}-x\right)\right]\)
\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\dfrac{x\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x^2+1\right)}\cdot\dfrac{1}{\left(x+1\right)^2\cdot\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
b) Thay \(x=-\dfrac{3}{2}\) vào C, ta được:
\(C=\dfrac{-3}{2}:\left(\dfrac{9}{4}+1\right)=\dfrac{-3}{2}:\dfrac{13}{4}=\dfrac{-3}{2}\cdot\dfrac{4}{13}=\dfrac{-6}{13}\)
c) Ta có: \(C=\dfrac{1}{2}\)
nên \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)(Loại)
Cho biểu thức C=\(\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\) Với \(x\ne\pm1\)
a, Rút gọn C
b, Tính giá trị của C khi \(x=-1\dfrac{1}{2}\)
c, Tìm x biết \(C=\dfrac{1}{2}\)
d, Chứng minh 2C<1
Rút gọn các biểu thức sau :
a) \(6x^n\left(x^2-1\right)+2x^3\left(3x^{n+1}+1\right)\)
b) \(3x^{n-2}\left(x^{n+2}-y^{n+2}\right)+y^{n+2}\left(3x^{n-2}-y^{n-2}\right)\)
c) \(x^{n-3}\left(x-y\right)+y\left(x^{n-3}+x^{n-3}y^{n-1}\right)\)
Rút gọn biểu thức:
\(A=\left|\frac{\left|y-x\right|}{\left|xy\right|}\right|+\left|\frac{y+x}{xy}-\frac{2}{z}\right|+\frac{\left|y-x\right|}{\left|xy\right|}+\frac{y+x}{xy}+\frac{2}{z}\)
với \(x>5\); \(y=\frac{x^2-25}{x+\frac{10x+25}{x}}\); \(z=\frac{x^2-25}{x+\frac{15x+25}{x-5}}\)
Rút gọn biểu thức:
\(\left(3x-2y\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)+\left(y+2x\right)^3\)
\(\left(3x-2y\right)^3+\left(y+2x\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)\)
\(=27x^3-54x^2y+36xy^2-8y^3+y^3+6xy^2+12x^2y+8x^3-\left(64x^3-125y^3\right)\)
\(=35x^3-42x^2y+42xy^2-7y^3-64x^3+125y^3\)
\(=-29x^3-42x^2y+42xy^2+118y^3\)
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.\(\left(x-5\right).\left(x+5\right)-\left(x+3\right)^2=2x-3\)
\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)
\(\Leftrightarrow-8x-31=0\)
\(\Leftrightarrow x=\dfrac{-31}{8}\)
\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-\left(x^3-5^3\right)=\left(x^3+2^3\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-x^3+5^3=x^3+2^3-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)
\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)
\(\Leftrightarrow96x=-117\)
\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)
2. \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2\)
\(\Leftrightarrow4x^2+12x+9+x^2-1=5\left(x^2+4x+4\right)\)
\(\Leftrightarrow4x^2+12x+9+x^2-1=5x^2+20x+20\)
\(\Leftrightarrow4x^2+x^2-5x^2+12x-20x=20-9+1\)
\(\Leftrightarrow-8x=12\)
\(\Leftrightarrow x=\dfrac{-12}{8}=\dfrac{-3}{2}\)