Tìm x
/x/ + / x+1 / + / x+4 / +....+ | x + 20| -11x = 0
| x - 7 | + | y + 11 | = 2
|6-x| + |y+2| = 1
| x+1| + / x+2 / + / x+3 / +....+ | x + 10| =11x
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1) Ta có : 2x2 + 3x - 5
= 2x2 - 2x + 5x - 5
= 2x(x - 1) + 5(x - 1)
= (x - 1) (2x + 5)
3) x2 + x - 6
= x2 + 2x - 3x - 6
= x(x + 2) - (3x + 6)
= x(x + 2) - 3(x + 2)
= (x - 3)(x + 2)
a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))
\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)
\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)
\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)
\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)
b) Thay x = 2003 ta có:
\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)
c) \(C>0\) khi:
\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\) (đpcm)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
a: 11x+4=-3/2
=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)
=>\(x=-\dfrac{1}{2}\)
b: \(x^2-9+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3+2\right)=0\)
=>(x-3)(x+5)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)
=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)
=>\(3x-9+10x+5=90\)
=>13x-4=90
=>13x=94
=>\(x=\dfrac{94}{13}\)
d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))
=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)
=>3x-11=2x-4-x-1
=>3x-11=x-5
=>2x=6
=>x=3(nhận)
f) 11.(x+2)=5.x+40
=> 11x+22=5x+40
=> 6x=18
=> x=18:6=3
g)11.(x-6)=4.x+11
=> 11.x+(-66)=4x+11
=> 7x=77
=> x=77:7=11
h) (3x-4):x+10=12
=> 3x:x-4:x=12-10=2
=> 3-4:x=2
=> 4:x=3-2=1
=> x=4:1=4
k)(5x+7):x+20=26
=> 5x:x+7:x=26-20=6
=> 5+7:x=6
=> => 7:x=6-5=1
=> x=7:1=7
m) x.1999-x=1999.1997+1999
=> x.1998=1999.(1997+1)
=> x.1998=1999.1998
=> x=1999
**** cho mik nha Nguyễn Thị Thảo Ly
Đổi x thành a, ta có:
f. 11 x ( a + 2) = 5 x a + 40
11 x a + 11 x 2 = 5 x a + 40
11 x a + 22 = 5 x a + 40
6 x a + 5 x a + 22 = 5 x a + 40
6 x a + 22 = 40(Cùng bỏ 5 x a ở hai vế)
6 x a = 40 - 22
6 x a = 18
a = 18 : 6
a = 3
g. 11 x ( a - 6 ) = 4 x a + 11
11 x a - 11 x 6 = 4 x a + 11
4 x a + 7 x a - 66 = 4 x a + 11
7 x a - 66 = 11(Cùng bỏ đi 4 x a ở 2 vế)
7 x a = 11 + 66
7 x a = 77
a = 77 : 7
a = 11
h. ( 3 x a - 4 ) : a + 10 = 12
3 x a - 4 = ( 12 - 10 ) x a
3 x a - 4 = 2 x a
2 x a + a = 2 x a + 4
a = 4(Cùng bỏ đi 2 x a)
k. Tương tự h
m. a x 1999 - a = 1999 x 1997 + 1999
a x 1998 = 1999 x 1998
=> a = 1999
\(x+\dfrac{2}{3}=\dfrac{9}{11}\Rightarrow x=\dfrac{9}{11}-\dfrac{2}{3}=\dfrac{5}{33}\\ x-\dfrac{3}{10}=\dfrac{4}{15}\Rightarrow x=\dfrac{4}{15}+\dfrac{3}{10}=\dfrac{17}{30}\\ x\times\dfrac{1}{7}=\dfrac{5}{6}\Rightarrow x=\dfrac{5}{6}:\dfrac{1}{7}=\dfrac{35}{6}\\ x:\dfrac{3}{5}=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{6}\times\dfrac{3}{5}=\dfrac{1}{10}\)