-(30-2x)-(55+6x)= 45+x
Bài 1: Tìm x
a) 5x - 7 = -19 + 2x
b) 35- 3x= 45 - 4x
c) 4x - 15 = - 75 - x
d) x - 12 = ( - 9 ) - 15
Bài 2 : Tìm x
a) 11- ( 15 + 14 ) = x - ( 25 - 9 )
b) 4x - 6 = - 20 + 3x
c) 45 - 2x = 55 - 5x
d) 6x - 15 =- 55 - 2x
Tìm x :
a. x x 45 + x x 55 = 1000
b. 6 x + 1 2 = 2
a. x x 45 + x x 55 = 1000
x x (45 + 55) = 1000
x x 100 = 1000
x = 1000 : 100
x = 10
b.
6 x + 1 2 = 2 6 : x + 1 2 = 2 6 : x = 2 - 1 2 6 : x = 3 2 x = 6 : 3 2 x = 4
Tìm x biết
a)(x+3)^2(x-2)^2=2x b)7x(x-2)=(x-2) c)8x^3-12x^2+6x-1=0
d)4x^2-9-x(2x-3)=0 e)x^3+5x^2+9x=-45 f)x^3-6x^2-x+30=0
d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
f) \(x^3-6x^2-x+30=0\)
\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)
\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)
Giải các phương trình sau :
a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
c, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\left(\dfrac{59-x}{49}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{45}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right).\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
Mà \(\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3=-7x+6x^2\)
\(\Leftrightarrow6x^2-5x+3+7x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy \(S=\left\{\dfrac{-3}{2}\right\}\)
c,\(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
\(\Leftrightarrow\dfrac{10x-40}{20}-\dfrac{6+4x}{20}=\dfrac{20x}{20}+\dfrac{4-4x}{20}\)
\(\Leftrightarrow\dfrac{6x-46}{20}=\dfrac{16x+4}{20}\)
\(\Leftrightarrow6x-46=16x+4\)
\(\Leftrightarrow6x-46-16x-4=0\)
\(\Leftrightarrow-10x-50=0\)
\(\Leftrightarrow-10x=50\)
\(\Leftrightarrow x=-5\)
Vậy \(S=\left\{-5\right\}\)
45 + 6 x 17 + 55 + 2x 3 x 83
45 + 6 x 17 + 55 + 2 x 3 x 83
= 45 + 102 + 55 + 6 x 38
= 45 + 102 + 55 + 228
= 147 + 55 + 228
= 202 + 228
= 430
tìm x biết
(x+3)^2-(x-2)^2=2x
b> 7x(x-2)=(x-2)
c> 8x^3-12x^2+6x-1=0
d>4x^2-9-x(2x-3)=0
e>x^3+5x^2+9x=-45
f>x^3-6x^2-x+30=0
g> x^2+16=10x
tìm x biết
a) 7x(x-2)=(x-2)
b) 4x^2-9-x(2x-3)=0
c) x^3+5x^2+9x=-45
d) x^3-6x^2-x+30=0
e) x^2+16=10x
a)\(7x\left(x-2\right)=\left(x-2\right)\)
\(\Leftrightarrow7x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7x-1=0\\x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=2\end{matrix}\right.\)
b)\(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)-x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)
c)\(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+9x+5x^2+45=0\)
\(\Leftrightarrow x\left(x^2+9\right)+5\left(x^2+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+9\right)=0\)
Dễ thấy: \(x^2+9\ge 9 >0\forall x\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
d,e tương tự
1+2+...+x=55
2+4+...+2x=2010
6x+4x=2010
x-55/45+x-30/35+x-25/25+x-40/15=10
các bãn giải được không ạ
\(\dfrac{x-55}{45}+\dfrac{x-30}{35}+\dfrac{x-25}{25}+\dfrac{x-40}{15}=10\)
\(< =>\dfrac{x-55}{45}+\dfrac{x-30}{35}+\dfrac{x-25}{25}+\dfrac{x-40}{15}-10=0\)
\(< =>\dfrac{x-55}{45}-1+\dfrac{x-30}{35}-2+\dfrac{x-25}{25}-3+\dfrac{x-40}{15}-4=0\)
\(< =>\dfrac{x-100}{45}+\dfrac{x-100}{35}+\dfrac{x-100}{25}+\dfrac{x-100}{15}=0\)
\(< =>\left(x-100\right)\left(\dfrac{1}{45}+\dfrac{1}{35}+\dfrac{1}{25}+\dfrac{1}{15}\right)=0\)
\(< =>x-100=0\left(\dfrac{1}{45}+\dfrac{1}{35}+\dfrac{1}{25}+\dfrac{1}{15}\ne0\right)\)
\(< =>x=100\)