2^x-1+2^x+3=68
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1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)
2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)
3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)
\(=7x^2-24x+13\)
4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)
\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)
5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)
6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)
7) \(\left(x+2\right)\left(x^2-2x+4\right)-x^2\left(x-2\right)-2x^2=x^3+8-x^3+2x^2-2x^2=8\)
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minh ko biết toán lớp 8 mik lớp 4 mà giải hộ mik với :(
68 x 5 + 68 x 5, 444 + 556 + 6825 + 2175
12 739 - (7000 - 739)
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TL:
\(x.\left(x+1\right).\left(x+2\right).\left(x+3\right)\rightarrow x=-2\)
\(\Rightarrow-2.\left(-2+1\right).\left(-2+2\right).\left(-2+3\right)\)
\(=-2.\left(-1\right).0.1\)
\(=0\)
HT
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Ta có: ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + .... + ( x + 9 ) + ( x + 10 ) = 240
<=> ( x + x + x + ...+ x ) + ( 1 + 2 + 3 + ... +10 ) = 240
<=> 10x + 55 = 240
=> 55 + 10x = 240
=> 10x = 240 - 55
=> 10x = 185
=> x = 185 : 10 = 18,5
Vậy x = 18.5
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a, \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x\left(x-3\right)=10\)
\(\Leftrightarrow x^3-6x^2+12x-8-x\left(x^2-1\right)+6x^2-18x=10\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x=10\)
\(\Leftrightarrow-5x=18\)
\(\Leftrightarrow x=\frac{-18}{5}\)
b, \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow\left[x+1-\left(x-1\right)\right]\left[\left(x+1\right)^2+x^2-1\left(x-1\right)^2\right]-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left(x^2+2x+1+x^2-1+x^2-2x+1\right)-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow2\left(3x^2+1\right)-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\frac{-6}{12}=\frac{-1}{2}\)
c,\(x^3+3x^2+3x+28=0\)
\(\Leftrightarrow\left(x+1\right)^3+27=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
\(\Leftrightarrow x=-4\)
d, \(x^3-6x^2+12x-7=0\)
\(\Leftrightarrow\left(x-2\right)^3+1=0\)
\(\Leftrightarrow\left(x-2\right)^3=-1\)
\(\Leftrightarrow x-2=-1\)
\(\Leftrightarrow x=1\)
a) (x - 2)3 - x (x + 1) (x - 1) + 6x (x - 3) = 10
x3 - 6x2 + 12x - 8 - x (x2 - 1) + 6x2 - 18x = 10
x3 - 6x2 + 12x - 8 - x3 + x + 6x2 - 18x = 10
-5x = 10 + 8
x = -3,6
b) (x + 1)3 - (x - 1)3 - 6 (x - 1)2 = -10
x3 + 3x2 + 3x + 1 - (x3 - 3x2 + 3x - 1) - 6 (x2 - 2x + 1) = -10
x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 + 12x - 6 = -10
12x - 4 = -10
12x = -6
x = -0,5
c) x3 + 3x2 + 3x + 28 = 0
(x + 1)3 + 27 = 0
(x + 1)3 = -27
(x + 1)3 = (-3)3
=> x + 1 = -3 => x = -4
d) x3 - 6x2 + 12x - 7 = 0
x3 - 6x2 + 12x - 8 + 1 = 0
(x - 2)3 = -1
(x - 2)3 = (-1)3
=> x - 2 = -1 => x = 1
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ĐKXĐ: x khác -2;-1;0;1.
\(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{3x-3}=\frac{1}{5x}\)
\((\frac{1}{x+1}-\frac{1}{5x})+(\frac{1}{x+2}+\frac{1}{3x-3})=0\)
\(\frac{4x-1}{5x(x+1)}+\frac{4x-1}{(x+2)(3x-3)}=0\)
hoặc \(4x-1=0\) hoặc \(5x(x+1)=(x+2)(3x-3)\)
Phương trình thứ nhất có nghiệm x=0,25 (t/m đkxđ)
Phương trình thứ 2 vô nghiệm.
Vậy pt có tập nghiệm S={0,25}.
Chúc bạn học tốt!
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5/6 x 1/7 - 1/7 x 1/6 = \(\dfrac{1}{7}x\left(\dfrac{5}{6}-\dfrac{1}{6}\right)=\dfrac{1}{7}x\dfrac{4}{6}=\dfrac{2}{21}\)
8/15 : 4/5 x 3/2= \(\dfrac{8}{15}x\dfrac{5}{4}x\dfrac{3}{2}=\dfrac{120}{120}=1\)
2x-1+2x+3=68
2x ×1/21 +2x ×23 =68
2x ×(1/2 +8)=68 (tính chất phân phối )
2x × 17/2 =68
2x =68÷17/2
2x =8
X= 3
Đúng thì **** động viên mình nha*^O^*