tìm x biết
a,|(x+1/2)|2x-3/4||=2x-3/4
Bài 1 : Tìm thương Q và dư R sao cho A= B.Q+R biết ;
a) A = \(x^4+3x^3+2x^2-x-4\) và B = \(x^2-2x+3\)
b) A = \(2x^3-3x^2+6x-4\) và B = \(x^2-x+3\)
c) A = \(2x^4+x^3+3x^2+4x+9\) và B = \(x^2+1\)
d) A = \(2x^3-11x^2+19x-6\) và B = \(x^2-3x+1\)
c) A= \(2x^4-x^3-x^2-x+1\) và B = \(x^2+1\)
5: Tìm x biết a) x/3 =4/12 b) x-1/ x-2=3/5 c) 2x :6=1/4 d) x² +x/2x²+1=1/2
a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
Tìm x biết:
a; 3x.(2x+3)-(2x+5x).(3x-2)=8
b;4.(x-1)-3.(x^2-5)_x^2=(x-3)-(x+4)
c; 2.(3x-1).(2x+5)-6.(2x-1).(x+2)=-6
d; 3.(2x-1).(3x-1)-(2x-3.(9x-1)-3=3
Câu 3: Tìm x ∈ N, biết:
a) 3 x . 3 = 243 b) 2 x . 162 = 1024 c) 64.4x = 168 d) 2 x = 16Câu 4 : Tìm x, biết. a) 2 x .4 = 128 b) (2x + 1)3 = 125 c) 2x – 2 6 = 6 d) 49.7x = 24013:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
Tìm x, biết
a)-2x+ -3/4 = x-3/5
b)|1/2x-1/3|-3/2=1/4
\(-2x-\frac{3}{4}=x-\frac{3}{5}\)
\(3x=\frac{3}{5}-\frac{3}{4}=-\frac{3}{20}\)
\(x=-\frac{1}{20}\)
b) \(\left|\frac{x}{2}-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
=> \(\orbr{\begin{cases}\frac{x}{2}-\frac{1}{3}=\frac{7}{4}\\\frac{x}{2}-\frac{1}{3}=-\frac{7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}}\)
a, \(-2x-\frac{3}{4}=x-\frac{3}{5}\)
\(-2x-\frac{3}{4}-x+\frac{3}{5}=0\)
\(-3x-\frac{3}{20}=0\)
\(\frac{3}{20}=-3x\Leftrightarrow x=\frac{1}{20}\)
\(a,-2x+\frac{-3}{4}=x-\frac{3}{5}\)
\(-2x-x=\frac{3}{4}-\frac{3}{5}\)
\(-3x=\frac{15}{20}-\frac{12}{20}\)
\(-3x=\frac{3}{20}\)
\(x=\frac{3}{20}\div\left(-3\right)\)
\(x=\frac{3}{20}.\frac{-1}{3}\)
\(x=-\frac{1}{20}\)
Vậy \(x=-\frac{1}{20}\).
\(b,\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{6}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{25}{12}\\\frac{1}{2}x=-\frac{17}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}\)
Vậy \(x\in\left\{\frac{25}{6};-\frac{17}{6}\right\}\).
Tìm x, biết:
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)
\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)
\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)
\(\Rightarrow-14x-60=0\)
\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)
\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)
\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)
\(\Rightarrow-x^2+12x-32=7x-x^2-10\)
\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)
\(\Rightarrow5x-22=0\)
\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
15x + 25 - 8x + 12 = 5x + 16x + 96 + 1
15x - 8x - 5x - 16x = 96 + 1 - 25 - 12
-14x = 60
x = \(\frac{60}{-14}\)
x = \(-\frac{30}{7}\)
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x2 + 2x
(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x2 + 2x
(x - 4)(-x + 8) = 5x - 10 - x2 + 2x
-x2 + 8x + 4x - 32 = 5x - 10 - x2 + 2x
(-x2 + x2) + (8x + 4x - 5x - 2x) = -10 + 32
5x = 22
x = \(\frac{22}{5}\)
a) 5(3x+5)-4(2x-3)=5x+8(2x+12)+1
<=> 15x+25-8x+12=5x+16x+96+1
<=>15x-8x-5x-16x=-25-12+96+1
<=> -14x = 60
<=> x= \(\frac{-60}{14}\)=\(\frac{-30}{7}\)
Vậy x= \(\frac{-30}{7}\)
Tìm x biết:
a)(2x-3)-(x-5)=(x+7)-(x+2)
b)(7x-5)-(6x+4)+(2x+3)-(2x+1)
c)(9x-3)-(8x+5)=(3x+2)
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
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Cho f(x)= x^5-2x^2+x^4 -x^5+3x^2-x^4-3+2x
a, Tìm x để f(x)= -3
b, Tìm nghiệm của f(x) biết :x^2+2x-3=(x-1).(x+3)
Giúp mik với!!!!!!!!
a) Ta có: f(x)=-3
<=>x5-2x2+x4-x5+3x2-x4-3+2x=-3
<=>(x5-x5)+(-2x2+3x2)+(x4-x4)+2x-3=-3
<=>x2+2x-3=-3
<=>x2+2x=0
<=>x(x+2)=0
<=>x=0 hoặc x+2=0
<=>x=0 hoặc x=-2
Vậy..........
b)đa thức f(x) có nghiệm
<=>f(x)=0
<=>x2+2x-3=0
<=>x2+3x-x-3=0
<=>x(x+3)-(x+3)=0
<=>(x-1)(x+3)=0
<=>x-1=0 hoặc x+3=0
<=>x=1 hoặc x=-3
Vậy nghiệm của đa thức f(x) là x=-3;x=1
Cho f(x)= x^5-2x^2+x^4 -x^5+3x^2-x^4-3+2x
a, Tìm x để f(x)= -3
b, Tìm nghiệm của f(x) biết :x^2+2x-3=(x-1).(x+3)
Giúp mik với!!!!!!!!