CMR:
\(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6}+\frac{x}{6-x}=-1\)
Giúp mk vs ạ!!!
\(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{6}{6-x}\)
hãy rút gọn biểu thức trên
giúp mk vs . :'(
= ( x/(x-6)(x+6) - x-6/x(x+6) ) : 2x-6/x2 + 6x + 6/6-x
=( x2/x(x+6)(x-6) - (x -6 )(x-6)/x(x+6)(x-6) ) : .....
= (12x -36 / x(x+6)(x-6) : 2x-6/ x2 + 6x )+ 6/6-x
=6/x-6 + 6/6-x
= 6-6/ x-6
=0/x-6
câu trước mình thiếu 6/6-x
câu trên sai , để mình giải đàng hoàng ra
Rút gọn : \(\left(\frac{x}{x^2-36}+\frac{6-x}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
\(\left(\frac{x}{x^2-36}+\frac{6-x}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
đkxđ: \(x\ne0;x\ne\pm6\)
MTC=x(x+6)(x-6)
\(=\left[\frac{x}{\left(x+6\right)\left(x-6\right)}+\frac{6-x}{x\left(x+6\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)
\(=\left[\frac{x^2}{x\left(x^2-36\right)}-\frac{\left(x-6\right)^2}{x\left(x^2-36\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)
\(=\frac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)
\(=\frac{12}{x\left(x-6\right)}-\frac{x^2}{x\left(x-6\right)}\)
\(=\frac{12-x^2}{x\left(x-6\right)}\)
.....................
Rút gọn : A = \(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right)\div\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
A = \(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
= \(\left[\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\left[\frac{x^2}{x\left(x-6\right)\left(x+6\right)}-\frac{\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{\left(x-x+6\right)\left(x+x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
=
= \(\frac{x\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2x-6}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{2x-6}{\left(x-6\right)\left(x+6\right)}.\frac{x\left(x+6\right)}{2x-6}\) \(-\frac{x}{x-6}\)
= \(\frac{x}{x-6}-\frac{x}{x-6}\)
= 0
Chứng minh rằng
a, \(\left(\frac{x}{x-36}-\frac{x-6}{x^2-6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}=-1\)
help meeee !!!
\(\left(\frac{x}{^{x^2-36}}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
chứng minh biểu thức không phụ thuộc vào biến x
S=\(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
a, Rút gọn biểu thức S
b, tìm x để giá trị của S=-1
\(a,\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(b,\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(c,\left(-3\right)^{x+5}=\frac{1}{81}\)
\(d,\left(\frac{1}{9}^x\right)=\left(\frac{1}{27}\right)^6\)
\(e,\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(f,5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(r,4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(h,\left(\frac{1}{2}-\frac{1}{3}\right).6x+6^{x+2}=6^{10}+6^7\)
nhờ mấy bn giúp mk tối mình nộp rồi
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
f)\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
r)\(4\cdot3^{x-1}+2\cdot3^{x+2}=4\cdot3^6+2\cdot3^9\)
\(4\cdot3^x:3+2\cdot3^x\cdot9=4.3^7:3+2\cdot3^7\cdot9\)
\(3^x\left(4:3+2\cdot9\right)=3^7\left(4:3+2\cdot9\right)\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
tìm điều kiện xác định của x để giá trị của biểu thức xác định và chứng minh rằng với điều kiện đó biểu thức không phụ thuộc vào biến
a. \(\left(x-\frac{1}{x}\right):\left(\frac{x^2+2x+1}{x}-\frac{2x+2}{x}\right)\)
b. \(\left(\frac{x}{x+1}+\frac{1}{x-1}\right):\left(\frac{2x+2}{x-1}-\frac{4x}{x^2-1}\right)\)
c. \(\frac{1}{x-1}-\frac{x^{3-x}}{^{x^2+1}}.\left(\frac{x}{x^2-2x+1}-\frac{1}{x^2-1}\right)\)
d. \(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
Tìm x :\(\left(\frac{x^2}{x^2-36}-\frac{x+6}{x^2-6x}\right):\frac{2x+6}{x^2-6x}-\frac{x}{x+6}\)
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