a ) \(\left(x-1\right)\left(x+1\right)-x\left(x+1\right)=35\)
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AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
AH
Akai Haruma
Giáo viên
30 tháng 4 2022
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
QA
1
21 tháng 8 2016
a ) \(\left(x-1\right)\left(x+1\right)-x\left(x+1\right)=35\)
\(\Leftrightarrow x^2-1^2-x.x-x.1=35\)
\(\Leftrightarrow x^2-1^2-x^2-x=35\)
\(\Leftrightarrow-1-x=35\)
\(\Leftrightarrow-x=35+1\)
\(\Leftrightarrow-x=36\)
\(\Leftrightarrow x=-36\)
QD
13 tháng 7 2017
a) \(-2x\left(10x-3\right)+5x\left(4x+1\right)=25\)
\(-20x^2+6x+20x^2+5x=25\)
\(\Rightarrow6x+5x=25\)
\(\Rightarrow11x=25\)
\(\Rightarrow x=\dfrac{25}{11}\)
b) \(y\left(5-2y\right)+2y\left(y-1\right)=15\)
\(5y-2y^2+2y^2-2y=15\)
\(\Rightarrow5y-2y=15\)
\(\Rightarrow3y=15\)
\(\Rightarrow y=5\)
c)\(x\left(x+1\right)-\left(x+1\right)=35\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=35\)
\(\Rightarrow x^2-1=35\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6;x=-6\)
d)\(x\left(x^2+x+1\right)-x^2\left(x+1\right)=0\)
\(x^3+x^2+x-x^3+x=0\)
\(\Rightarrow x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x=0;x=0-2=-2\)
Vậy \(x=0;x=-2\)
25 tháng 5 2022
a: Sửa đề: \(\left(x-18\right)\left(x-7\right)\left(x+35\right)\left(x+90\right)-67x^2\)
\(=\left(x^2+17x-630\right)\left(x^2+83x-630\right)-67x^2\)
Đặt \(x^2+17x-630=a\)
Ta sẽ được \(a\left(a+66x\right)-67x^2\)
\(=a^2+66ax-67x^2\)
\(=\left(a+67x\right)\left(a-x\right)\)
\(=\left(x^2+84x-630\right)\left(x^2+16x-630\right)\)
b: \(\left(3x-2\right)^3+\left(1-2x\right)^3+\left(1-x\right)^3\)
Đặt 1-2x=a; 1-x=b
Ta sẽ được \(-\left(a+b\right)^3+a^3+b^3\)
\(=-a^3-b^3-3ab\left(a+b\right)+a^3+b^3\)
\(=-3ab\left(a+b\right)\)
\(=-\left(2-3x\right)\left(1-2x\right)\left(1-x\right)\)
11 tháng 3 2017
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
AT
11 tháng 3 2017
a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)
28 tháng 5 2017
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
a ) \(\left(x-1\right)\left(x+1\right)-x\left(x+1\right)=35\)
\(x^2-1^2-x^2-x=35\)
\(-1-x=35\)
\(-x=35+1\)
\(-x=36\)
\(x=-36\)