CY
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K
Khách
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7 tháng 7 2019
Đề bài là: \(\frac{3\text{x}+5}{x^2-5\text{x}+25}-\frac{x}{25-5\text{x}}\)
hay: \(\frac{3\text{x}+5}{\frac{x^2-5\text{x}+25-x}{25-5\text{x}}}\)
thế bạn? 
8 tháng 12 2021
\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)
8 tháng 12 2021
\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)
\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)
\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)
\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)
\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)
\(=\dfrac{-\left(5-x\right)^2}{5x\left(5-x\right)}\)
\(=\dfrac{-5+x}{5x}\)
\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)
\(=\dfrac{x+1}{x+3}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-2x+x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+x-6}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)
\(=\dfrac{x^2+x-6}{x^2+x-6}\)
\(=1\)
PL
1
17 tháng 3 2020
a) (x-17)/33 + (x-21)/29 + x/25 = 4
<=> [(x-17)/33-1] + [(x-21)/29-1] + x/25-2] = 0
<=> (x-50)/33 + (x-50)/29 + (x-50)/25 = 0
<=> (x-50)(1/33+1/29+1/25) = 0
Mà 1/33+1/29+1/25 khác 0.
<=> x- 50 = 0 <=> x=50
b) (3x−5)(7−5x)+(5x+2)(3x−2)=2
<=> 21x−15x2−35+25x+15x2−10x+6x−4−2=0
<=> (15x2−15x2)+(25x+21x−10x+6x)−(35+4+2)=0
<=> 42x=41
<=> x=41/42
CL
1
13 tháng 9 2021
1) \(\left(5x-1\right)\left(5x+1\right)=25x^2-7x+15\)
\(\Leftrightarrow25x^2-1=25x^2-7x+15\)
\(\Leftrightarrow7x=16\Leftrightarrow x=\dfrac{16}{7}\)
2) \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow3x^2-2x-5-3x^2-2x+1=x-4\)
\(\Leftrightarrow5x=0\Leftrightarrow x=0\)
PQ
2
MH
2 tháng 12 2015
\(\left(\frac{3x-5}{x^2-5x}-\frac{x+5}{5x-25}\right):\frac{x^2-25}{x}\)
\(=\left[\frac{3x-5}{x\left(x-5\right)}-\frac{x+5}{5\left(x-5\right)}\right].\frac{x}{x^2-25}\)
\(=\left[\frac{\left(3x-5\right).5}{x\left(x-5\right).5}-\frac{\left(x+5\right).x}{5\left(x-5\right).x}\right].\frac{x}{x^2-25}\)
\(=\left[\frac{15x-25}{5x\left(x-5\right)}-\frac{x^2+5x}{5x\left(x-5\right)}\right].\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{15x-25-x^2-5x}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-x^2+10x-25}{5x\left(x-5\right)}.\frac{x}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-\left(x-5\right)^2.x}{5x\left(x-5\right)\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-1}{5\left(x+5\right)}\).
12 tháng 6 2018
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
17 tháng 3 2020
a) Ta có: \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)
\(\Leftrightarrow\frac{x-17-33}{33}+\frac{x-21-29}{29}+\frac{x-2\cdot25}{25}=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)
nên x-50=0
hay x=50
Vậy: x=50
b) Ta có: \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)=2\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-4x-4-2=0\)
\(\Leftrightarrow42x-41=0\)
\(\Leftrightarrow42x=41\)
hay \(x=\frac{41}{42}\)
17 tháng 3 2020
a, \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=4-4\)
\(\Leftrightarrow\left(\frac{x-17-33}{33}\right)+\left(\frac{x-21-29}{29}\right)+\left(\frac{x-2.25}{25}\right)=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\) (*)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\Rightarrow\) Phương trình (*) xảy ra khi: \(x-50=0\Leftrightarrow x=50\)
Vậy phương trình có nghiệm duy nhất là x = 50.
28 tháng 8 2019
a) = (x+3).(x-3)^2-(x-3)(x+3)^2
=(x^2-9)(x-3)-(x^2-9)(x+3)
=(x^2-9)(x-3-x-3)
=-6(x^2-9)
các câu còn lại tương tự
28 tháng 8 2019
\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+3-\left(x^3-3\right)\)
\(=x^3+3-x^3+3\)
\(=6\)
\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)
\(=x^3-5^3-x^3-5^3\)
\(=-125-125\)
\(=-250\)
=\(\frac{3x+5}{-x.\left(-x+5\right)}\)+\(\frac{25-x}{-5x+25}\)
=\(\frac{1x-25-x^2}{5x.\left(-\left(x-5\right)\right)}\)
=\(\frac{-\left(x^2-10x+25\right)}{5x.\left(-\left(x-5\right)\right)}\)
=\(\frac{x-5}{5x}\)