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\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Rightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
\(\Rightarrow x-100=0\).Do \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\ne0\)
\(\Rightarrow x=100\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{x-45}{55}-1-\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
\(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
(x-100)(\(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}=0\)
-> x-100 = 0 -> x = 100
mà \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\) khác 0
Vậy x = 100
x^3 - x^2 - 21x + 45 = 0
=>x^3 + 5x^2 - 6x^2 - 30x + 9x + 45 = 0
=> x^2(x + 5) - 6x(x + 5) + 9(x + 5) = 0
=> (x^2 - 6x + 9)(x + 5) = 0
=> (x - 3)^2(x + 5) = 0
=> x - 3 = 0 hoặc x + 5 = 0
=> x = 3 hoặc x = -5
Ta có: x3−x2+x−1=0
⇔x2(x−1)+(x−1)=0
⇔(x−1)(x2+1)=0(1)
Ta có: x2≥0∀x
⇒x2+1≥1≠0∀x(2)
Từ (1) và (2) suy ra x−1=0
⇔x=1Ta có: x3−x2+x−1=0
⇔x2(x−1)+(x−1)=0
⇔(x−1)(x2+1)=0(1)
Ta có: x2≥0∀x
⇒x2+1≥1≠0∀x(2)
Từ (1) và (2) suy ra x−1=0
⇔x=1
dễ thôi mà
Áp dụng tỉ lệ thức, ta có:
\(\Leftrightarrow\frac{108x-4970}{2915}=\frac{92x-4970}{2115}\Rightarrow\left(108x-4970\right)2115=2915\left(92x-4970\right)\)
=>x=100
ta có: (59-x)/41 +(57-x)/43 +(55-x)/45 +(53-x)/47 +(51-x)/49 =-5
<=>[(59-x)/41 +1 ] +[(57-x)/43 +1] +[(55-x)/45 +1] +[(53-x)/47 +1] +[(51-x)/49 +1] =0
<=>(59-x-41)/41 + (57-x-43)/43 +(55-x-45)/45 +(53-x-47)/47 +(51-x-49)/49 =0
<=>(100-x)/41 + (100-x)/43 + (100-x)/45 +(100-x)/47 + (100-x)/49 =0
<=>(100-x).( 1/41 + 1/43 + 1/45 + 1/47 + 1/49 ) =0
mà (1/41 + 1/43 + 1/45 + 1/47 + 1/49) khác 0 nên 100-x =0 <=>x=100
vậy nghiệm của pt là x=100
\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\left(\dfrac{x-45}{55}-1\right)+\left(\dfrac{x-47}{53}-1\right)=\left(\dfrac{x-55}{45}-1\right)+\left(\dfrac{x-53}{47}-1\right)\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
Do \(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\) nên x - 100 = 0 <=> x = 100
\(\frac{x+43}{57}+\frac{x+46}{54}+\frac{x+49}{51}+\frac{x+235}{45}=0\)
\(\Leftrightarrow\text{}\text{}\)\(\frac{x+43}{57}+1+\frac{x+46}{54}+1+\frac{x+49}{51}+1+\frac{x+235}{45}-3=0\)
\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}+\frac{x+100}{51}+\frac{x+100}{45}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}+\frac{1}{51}+\frac{1}{45}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy x = -100
ĐKXĐ : x khác 0 ; x khác 5
<=> \(\frac{45\left(x-5\right)}{x\left(x-5\right)}-\frac{45x}{x\left(x-5\right)}=\frac{3}{2}\)
<=> \(\frac{-225}{x\left(x-5\right)}=\frac{3}{2}\)
=> 3x( x - 5 ) = -450
<=> 3x2 - 15x + 450 = 0
<=> x2 - 5x + 150 = 0
Vì x2 - 5x + 150 = ( x - 5/2 )2 + 575/4 ≥ 575/4 ∀ x
nên pt vô nghiệm