1)tìm x
a)3(2-x)+5(x-6)=-98
b)(x^2+1)(49-x^2)=0
tìm x
a,(x+1)^3-(x-1)^3-6(x-1)^2=-10
b,x(x+5)(x-5)-(x+2)(x^2-2x+4)=42
c,(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=49
a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
a) 3(2 - x) + 5 ( x- 6) = - 98
b) ( x + 7) ( 8 - x ) = 0
a) `3(2-x)+5(x-6)=-98`
`6-3x+5x-30=-98`
`-3x+5x=-98-6+30`
`2x=-74`
`x=-37`
b) `(x+7)(8-x)=0`
\(\left[{}\begin{matrix}x+7=0\\8-x=0\end{matrix}\right.\\ \left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
a, \(3\left(2-x\right)+5\left(x-6\right)=-98\)
\(6-3x+5x-30=-98\)
\(2x-24=-98\)
\(2x=-74\)
\(x=-37\)
b,\(\left(x+7\right)\left(8-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
-Chúc bạn học tốt-
Giải:
a) \(3.\left(2-x\right)+5.\left(x-6\right)=-98\)
\(6-3x+5x-30=-98\)
\(-3x+5x=-98-6+30\)
\(2x=-74\)
\(x=-74:2\)
\(x=-37\)
b) \(\left(x+7\right).\left(8-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+7=0\\8-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
Chúc bạn học tốt!
Tìm x
a) -1/2 . x = 5
b) x-1/9 = 8/3
c) x + 5/6 = 16/42 - -8/56
d) 1 3/4x - 5 = -3 1/3
e) 3 1/3 - 3/4 : x = -1/6
g) 3/x + 5 = 15%
h) 1/2 + 1/2.3 + 1/3.4 +...+1/x(x+1) = 49/50
Tìm x
a) -1/2 . x = 5
b) x-1/9 = 8/3
c) x + 5/6 = 16/42 - -8/56
d) 1 3/4x - 5 = -3 1/3
e) 3 1/3 - 3/4 : x = -1/6
g) 3/x + 5 = 15%
h) 1/2 + 1/2.3 + 1/3.4 +...+1/x(x+1) = 49/50
a: x=5:(-1/2)=-10
b: x=8/3+1/9=25/9
c: =>x+5/6=11/21
=>x=-13/42
d: =>7/4x-5=-10/3
=>7/4x=5/3
=>x=20/21
e: =>10/3-3/4:x=-1/6
=>3/4:x=10/3+1/6=21/6=7/2
=>x=3/4:7/2=3/4*2/7=6/28=3/14
g: =>3/(x+5)=3/20
=>x+5=20
=>x=15
h: =>1-1/2+1/2-1/3+...+1/x-1/x+1=49/50
=>1-1/x+1=49/50
=>x+1=50
=>x=49
bài 7 tìm x
1,x(x+3)-5(x+3)=0 2,5x(x-1)=x-1
3,(x+1)=(x+1)\(^2\) 4,x(2x-3)-2(3-2x)=0
5,\(\left(x-2\right)^2-4=0\) 6,\(36x^2=49\)
7,\(2x\left(x-6\right)-x+6=0\) 8,\(3x\left(2x-1\right)-24x+12=0\)
9,\(x^2-6x+8=0\) 10,\(x^2+2x-15=0\)
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
tìm x
a) (x+2)(x+3)-(x-2)(x+5)=6
b) (3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)
c) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
Bài 1: Tìm x :
a. ( x + 4 )2 - ( x + 1 ) ( x - 1 ) = 16
b. ( 2x - 1 )2 + ( x + 3 )2 - 5 ( x + 7 ) ( x - 7 ) = 0
c. ( x - 2 )3 - ( x - 3 ) ( x2 + 3x + 9 ) + 6 ( x + 1 )2 = 49
Bài 1: Tìm x :
a. ( x + 4 )2 - ( x + 1 ) ( x - 1 ) = 16
b. ( 2x - 1 )2 + ( x + 3 )2 - 5 ( x + 7 ) ( x - 7 ) = 0
c. ( x - 2 )3 - ( x - 3 ) ( x2 + 3x + 9 ) + 6 ( x + 1 )2 = 49
bài 1 tìm x
a. 5 - 3(x+4) = -1
b.(x-1) - (x+2) = 0
c.( \(\dfrac{1}{2}\) + x )-( \(\dfrac{1}{3}\) - x) = 0
d. 2x2 - 3 = 5
e. x(2x -1) = 0
g. \(\dfrac{1}{3}\) . x2 - \(\dfrac{1}{6}\)=\(\dfrac{7}{6}\)
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
a)\(=>3\left(x+4\right)=6=>x+4=2=>x=-2\)
b)\(=>x-1-x-2=0\)
\(=>-3=0\left(vl\right)\) => x ko tồn tại