C=x+9/x-6
D=x+3/x+1
E=x-4/x-1
F=6-x/3-x
C=x+9/x-6
D=x+3/x+1
E=x-4/x-1
F=6-x/3-x
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)
\(x=0\)
\(b.x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}+\dfrac{1}{4}\)
\(x=\dfrac{-1}{4}\)
c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)
\(-2x=18\)
\(x=-9\)
d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)
\(4.\left(9-x\right)=-60\)
\(9-x=-15\)
\(x=24\)
\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)
\(\left(x+1\right)^2=9\)
\(\left[{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)
\(\left(x-1\right)^2=16\)
\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
a, x.2 + x.5 = 14 b, x.9 + x = 20
c, x: \(\dfrac{3}{2}\) + x.\(\dfrac{1}{3}\) = 1999 d,11.(x+2) = 5.x + 40
e, 11.(x-6) = 4.x + 11 f, (3.x - 4) : x + 10 = 12
g, (5.x + 7) : x + 20 = 26 h, x.1999 - x = 1999.1997 + 1999
***dấu chấm(.) = dấu nhân(x)***
a) \(...\Rightarrow x.\left(2+5\right)=14\Rightarrow x.7=14\Rightarrow x=14:7=2\)
b) \(...\Rightarrow x.\left(9+1\right)=20\Rightarrow x.10=20\Rightarrow x=20:10=2\)
c) \(...\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=1999\Rightarrow x.\dfrac{3}{3}=1999\Rightarrow x=1999\)
d) \(...\Rightarrow11.x+22=5.x+40\Rightarrow11.x-5.x=40-22\Rightarrow6.x=18\Rightarrow x=18:6=3\)
e) \(...\Rightarrow11.x-66=4.x+11\Rightarrow11.x-4.x=11+66\Rightarrow7.x=77\Rightarrow x=77:7=11\)
f) \(...\Rightarrow\left(3.x-12\right):x=12-10\)
\(\Rightarrow3.x-12=2.x\)
\(\Rightarrow3.x-2.x=12\)
\(\Rightarrow x=12\)
g) \(...\Rightarrow\left(5.x+7\right):x=26-20\)
\(\Rightarrow5.x+7=6.x\)
\(\Rightarrow6.x-5.x=7\)
\(\Rightarrow x=7\)
h) \(...\Rightarrow x.\left(1999-1\right)=1999.\left(1997+1\right)\)
\(\Rightarrow x.1998=1999.1998\)
\(\Rightarrow x=1999.1998:1998\)
\(\Rightarrow x=1999\)
a, \(x\times\) 2 + \(x\times\) 5 = 14
\(x\) \(\times\) ( 2 + 5) = 14
\(x\) \(\times\) 7 = 14
\(x\) = 14: 7
\(x\) = 2
b, \(x\times9\) + \(x\)= 20
\(x\) \(\times\)( 9 + 1) = 20
\(x\) \(\times\) 10 = 20
\(x\) = 2
c, \(x\) : \(\dfrac{3}{2}\) + \(x\times\dfrac{1}{3}\) = 1999
\(x\times\) \(\dfrac{2}{3}\) + \(x\) \(\times\dfrac{1}{3}\) = 1999
\(x\times\) ( \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) = 1999
\(x\) = 1999
d, 11\(\times\)(\(x+2\)) = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) + 22 = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 40 - 22
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 18
11 \(\times\) \(x\) - 5 \(\times\) \(x\) = 18
\(x\) \(\times\) ( 11 - 5) = 18
\(x\) \(\times\) 6 = 18
\(x\) = 3
e, 11 \(\times\)( \(x-6\)) = 4 \(\times\) \(x\) + 11
11 \(\times\)\(x\) - 66 = 4 \(\times\) \(x\) + 11
11 \(\times\) \(x\) = 4 \(\times\) \(x\) + 11 + 66
11 \(\times\) \(x\) = 4 \(\times\) \(x\) + 77
11 \(\times\) \(x\) - 4 \(\times\) \(x\) = 77
\(x\) \(\times\) ( 11 - 4) = 77
\(x\) \(\times\) 7 = 77
\(x\) = 77 : 7
\(x\) = 11
g, (5 \(\times\) \(x\) + 7) : \(x\) + 20 = 26
(5 \(\times\) \(x\) + 7) : \(x\) = 26 - 20
(5 \(\times\) \(x\) + 7) : \(x\) = 6
5 \(\times\) \(x\) + 7 = 6 \(\times\) \(x\)
6 \(\times\) \(x\) - 5 \(\times\)\(x\) = 7
\(x\times\)( 6- 5) = 7
\(x\) = 7
f, (3 \(\times\) \(x\) - 4): \(x\) + 10 = 12
(3 \(\times\) \(x\) - 4) : \(x\) = 12 - 10
( 3 \(\times\) \(x\) - 4): \(x\) = 2
3 \(\times\) \(x\) - 4 = 2 \(\times\) \(x\)
3 \(\times\) \(x\) - 2 \(\times\) \(x\) = 4
\(x\) \(\times\) ( 3 - 2) = 4
\(x\) = 4
h, \(x\times\) 1999 - \(x\) = 1999 \(\times\)1997 + 1999
\(x\) \(\times\)( 1999 - 1) = 1999 \(\times\) ( 1997 + 1)
\(x\) \(\times\) 1998 = 1999 \(\times\) 1998
\(x\) = 1999 \(\times\) 1998 : 1998
\(x\) = 1999
bài 6:Tìm x, biết:A,1/4x-1/3=-5/9;B,3,5-Ix-1/2I=0,75;C,x-1/x-5=6/7;D,(x-4)mũ 2=25;E,2 mũ x+2 mũ x-4=272;F,(x+1/2)(2/3-2x)=0
1.Tìm x, biết:
a) 7/8 + x = 4/7 b) 6 – x = (-3)/4 c) 1/(-5) + x = 3/4 d) –6 – x = (-3)/5
e) - 2/6 + x = 5/7 f) -8 – x = (-5)/3 g) - 2/10 – x = - 9/8 h) 8 – x = (-1)/6
i) - 2/5 + x = 5/7 j) -2 – x = (-5)/3 k) - 1/6 – x = - 9/8 l) 8 – x = (-1)/5
2. Tính:
a) -7/3 + 4/9 b) -8/18 - 15/56 c) (-7) - (-3/4) d) 0,15 + (-5/12)
e) -1/42 + -1/28 f) -1/21 + -1/36 g) 5,5 - (-2/7) h) 0,9 - (-4/7)
Phân tích các đa thức sau thành nhân tử:
a) \({\left( {x - 1} \right)^2} - 4\)
b) \(4{x^2} + 12x + 9\)
c) \({x^3} - 8{y^6}\)
d) \({x^5} - {x^3} - {x^2} + 1\)
e) \( - 4{x^3} + 4{x^2} + x - 1\)
f) \(8{x^3} + 12{x^2} + 6x + 1\)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
a,|x|-7/6=9/15
b,|x-4/3|=1/6
c,|x-4/3|-1/3=1/2
d,8/3-|7/9-x|=-1/5
e,|x-1/4^2|-25/64=0
f,(x-1/4)^2+17/64=21/32
a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
a, x.2 + x .5 = 14 b, x.9 + x = 20
c, x: \(\dfrac{3}{2}\) + x. \(\dfrac{1}{3}\) = 1999 d, 11.(x+2) = 5.x + 40
e, 11.(x-6) = 4.x + 11 f, (3.x - 4) : x + 10 = 12
g, (5.x + 7) : x + 20 = 26 h, x.1999 - x = 1999.1997 + 1999
A.(X+16/49) +(x+18/47)=(x+20/45)-1
B.(x-3/x-2)+(x-2/x-4)=-1
C.(4x-10).(x+6)=0
D.7x-1/2=5+(9-5x/6)
E.3.(7x-1)=30+9-5
F.(3x-9/x+1)-2=4x/x+1