Tìm x biết :
\(\left(x+2\right)^3-\left(x-2\right)^3-16=108\)
Tìm x, biết
a) \(\left(x+3\right)^4-\left(x-3\right)^4-24x^3=108\)
b) \(\left(x+2\right)^5-\left(x-2\right)^5=64\)
Tìm x, y biết :
\(\left|x+3\right|+\left|x-1\right|=\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\)
Ta có: \(\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)
\(\left|y-2\right|+\left|y+2\right|=\left|2-y\right|+\left|y+2\right|\ge\left|2-y+y+2\right|=4\)
\(\Rightarrow\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\le\dfrac{16}{4}=4\Rightarrow\left|x+3\right|+\left|x-1\right|\ge\dfrac{6}{\left|y-2\right|+\left|y+2\right|}\)
Dấu '=' xảy ra <=> (x+3)(1-x)\(\ge0\) và (2-y)(y+2)\(\ge0\)
Vì x,y \(\in Z\Rightarrow\left\{{}\begin{matrix}x\in\left\{-3;-2;-2;0;1\right\}\\y\in\left\{-2;-1;0;1;2\right\}\end{matrix}\right.\)
Tìm x:a, \(\left(x+3\right)^4-\left(x-3\right)^4-24x^2\left(x-1\right)=108\)
b, \(\left(x+2\right)^5-\left(x-2\right)^5=108\)
tìm x biết:
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
b, \(\left(2x-1\right)^2-\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
a,\((x+4)^2-(x+1)(x-1)=16\)
\(\Rightarrow x^2+8x+16-x^2+1=16\)
\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)
b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)
\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)
\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)
\(\Rightarrow -x^2-10x-244=0\)
\(\Rightarrow -(x^2-10x+25)-219=0\)
\(\Rightarrow -(x-5)^2-219=0\)
\(\Rightarrow (x-5)^2+219=0\)
Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm
Bài 1 Tìm x
\(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x^2-16\right)+3x^2\)
Tìm x biết \(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)
NX: 2x+3; 5(2x+3) và 2(2x+3) cùng dấu
+TH1: 2x+3 \(\ge\)0 => x \(\ge\frac{-3}{2}\)
=> 5(2x+3), 2(2x+3) \(\ge\)0
=> |5(2x+3)| = 5(2x+3); |2(2x+3)| = 2(2x+3); |2x+3| = 2x+3
=> (2x+3)(5+2+1) = 16
=> 2x+3 = 2
=> 2x = -1
=> x = -1/2 (t/m)
+ TH2: 2x+3 < 0 => x < -3/2
cmtt => -5(2x+3) - 2(2x+3) - (2x+3) = 16
=> (2x+3)(-5-2-1) = 16
=> 2x+3 = -2
=> 2x = -5
=> x = -5/2 (t/m)
/8(2x+3/ = 16
/2x+3/=2
2x+3=2 hoặc 2x+3=-2
2x=-1 hoặc 2x=-5
x=-1/2 hoặc x=-5/2
bạn trả lời nhé
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
Tìm x biết: \(\frac{4}{\left(x+2\right).\left(x+6\right)}+\frac{7}{\left(x+6\right).\left(x+13\right)}=\frac{2x+1}{\left(x+2\right).\left(x+16\right)}-\frac{3}{\left(x+13\right).\left(x+16\right)}\)
Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)
<=>1/x+2 -1/x+6 +1/x+6 -1/x+13
<=>1/x+2-1/x+13
=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)
<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)
<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)
<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)
<=> 2x+1=14
<=> 2x=14-1
<=> 2x=13
<=> x=13:2
<=> x=13/2
Vậy x=13/2
Chắc là vầy. Mk cug ko chắc nữa
Tìm x biết :
\(a)\)\(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+3\right|=16\)
\(b)\)\(\left|x^2+\left|6x-2\right|\right|=x^2+4\)