Để B là số nguyên thì x-5+8 chia hết cho x-5
=>8 chia hết cho x-5
\(\Leftrightarrow x-5\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
hay \(x\in\left\{6;4;7;3;9;1;13;-3\right\}\)
Để B là số nguyên thì x-5+8 chia hết cho x-5
=>8 chia hết cho x-5
\(\Leftrightarrow x-5\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
hay \(x\in\left\{6;4;7;3;9;1;13;-3\right\}\)
Bài 1. Tìm \(x\).
a) -5(\(x\)2-3\(x\)+1)+\(x \)(1+5\(x\))=\(x-2\)
b) \(12x\)2\(-4x\)\((3x+5)\)\(=10x-17\)
c) \(-4x(x-5)+7x(x-4)-3x\)2\(=12\)
Bài 2. Tính ( Rút gọn).
a) \((x+5).(x-7)-7x.(x-3)\)
b) \(x.(x\)2\(-x-2)-(x-5).(x+1)\)
c) \((x-5).(x-7)-.(x+4).(x-3)\)
d) \((x-1).(x-2)-(x+5).(x+2)\)
Cho đa thức B(x) = 2\(x^{^{ }2}\)-4x + 3. Tính B(3), B(-\(\dfrac{1}{2}\) )
Cho đa thức M(x) = 7\(x^3\)- 3\(x^4\)- \(x^2\) + 3\(x^2\)- \(x^3\)- 3\(x^4\)- 6\(x^3\)
Cho đa thức N(x) = 3x - 5\(x^3\) + 8\(x^2\)- 5x + 5\(x^3\) + 5
cho hai da thuc A(x)=2x(x-2)-5(x+3)+7x^3 va B(x)=-x(x+5)-(2x-3)+x(3x^2-2x).a, thu gon A(x),B(x).b, tim nghiem cua da thuc P(x)=A(x)-B(x)-x^2(4x+5)
a,5^x+4-3×5^x+3=2×5^11
b,3×5^x+2+4×5^x-3=19×5^10
c,6×8^x-1+8^x-1=6×8^19+8^21
d,5×2^x+3×2^x+2=5×2^5+3×2^7
a, 4x(x2-x-1)-(x2-2)(x+3)
b, (x+5)(x+7)-7x(x+3)
c, x(x2-x-2)-(x+5)(x-1)
d, (x+5)(x+7)-(x-4)(x+3)
Tìm x nguyên sao cho:
a) 5*3^x+6 = 2*3^5 + 3*3^5
b) 5^x+4-3*5^x+3=2*5^4
b) 3.(x−1/2)−5(x+3/5)=−x+1/5
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
1) Tìm x
a) -5.(x+1/5)-1/2.(x-2/3)=3/2.x-5/6
b) 3.(x-1/2)-5.(x+3/5)=-x+1/5
c) /x/ - 38/7./x/-3/4=2./x/+(-8/7)
/x/ là giá trị tuyệt đối nhé các bn
Bài 1: a, x + 0,5 = -2/3 b, x - (-2/5) = 5/7 c, 5/4 - 6x = 7 1/2 d, 3 - 2x - (5/4 - 7/5) = 9/20 bài 2: a, x : (-2/3) mũ 3 = -2/3 b, (2/5) mũ 5 nhân x = (2/5) mũ 7 c, 343/125 = (7/5) mũ x d, (-1/3) nhân x = 1/243