ai rảnh giúp mik vs l x-2 l +3 =2x
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a) x(x-5)+2(x-5) = (x-5)(x+2)
b) (x-7)(x-2)
c) (x+2)(x^2+2x+4)+5y(x+2) = (x+2)(x^2+2x+4+5y)
d) (x^2+8)^2 -16x^2 = (x^2+8-4x)(x^2+8+4x)
a) A(x) +B(x)=\(( x^4+2x^3-5x^2-3x-6)+(-x^4-2x^3+5x^2+x+10)\)
=\(x^4+2x^3-5x^2-3x-6+-x^4-2x^3+5x^2+x+10\)
=\((x^4 -x^4)+(2x^3 -2x^3 )-(5x^2-5x^2)-(3x-x)-(6-10)\)
=-2x +4
A(x)-B(x)=\( ( x^4+2x^3-5x^2-3x-6)-(-x^4-2x^3+5x^2+x+10)\)
=\(x^4+2x^3-5x^2-3x-6+x^4+2x^3-5x^2-x-10 \)
=\((x^4 +x^4)+(2x^3 +2x^3)-(5x^2 +5x^2)-(3x+x)-(6+10)\)
=\(2x^4+4x^3-10x^2-4x-16\)
b) B(x)-M(x)=A(x)
M(x)=B(x)-A(x)
=\((-x^4-2x^3+5x^2+x+10)-( x^4+2x^3-5x^2-3x-6)\)
=\(-x^4-2x^3+5x^2+x+10- x^4-2x^3+5x^2+3x+6\)
=\((-x^4 -x^4)-(2x^3+2x^3)+(5x^2 +5x^2)+(x+3x)+(10+6)\)
=\(-2x^4-4x^3+10x^2+4x+16\)
\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
a: \(\left(x-3\right)\left(2x^2-3x+4\right)\)
\(=2x^3-3x^2+4x-6x^2+9x-12\)
\(=2x^3-9x^2+13x-12\)
b: \(\left(4x^2y-5xy^2+6xy\right):2xy\)
\(=\dfrac{4x^2y-5xy^2+6xy}{2xy}\)
\(=\dfrac{2xy\cdot2x-2xy\cdot2,5y+2xy\cdot3}{2xy}\)
\(=2x-2,5y+3\)
c: \(\dfrac{x}{2x+4}-\dfrac{2}{x^3+2x}\)
\(=\dfrac{x\left(x^3+2x\right)-2\left(2x+4\right)}{x\left(x^2+2\right)\cdot2\cdot\left(x+2\right)}\)
\(=\dfrac{x^4+2x^2-4x-8}{2x\left(x^2+2\right)\left(x+2\right)}\)
nhìn có vẻ đề khá logic nhưng khi bn giải thì s '' thì nó khá đơn giản ''
\(\left(x-1\right)\left(x+5\right)-2\left(2x-1\right)=\left(x-2\right)3\)
\(x^2-x^2-5=3x-6\)
\(0-5-3x-6=0\)
\(-11-3x=0\)
\(3x=-11\)
\(x=-\frac{11}{3}\)
ns tớ sai đâu sửa nha !
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
\(\dfrac{11}{2}.\dfrac{21}{3}+\dfrac{11}{3}.\dfrac{1}{2}\)
\(=\dfrac{77}{2}+\dfrac{11}{6}=\dfrac{121}{3}\)
| x - 2 | + 3 = 2x
* | x - 3 | = x - 3 khi x - 3\(\ge\)0 hay x \(\ge\)3
| x - 3 | = -( x - 3 ) khi x - 3 < 0 hay x < 3
Quy về giải hai phương trình :
* x - 3 + 3 = 2x ( x \(\ge\)3 )
<=> x = 2x
<=> x - 2x = 0
<=> -x = 0
<=> x = 0 ( không tmđk )
* -( x - 2 ) + 3 = 2x ( x < 0 )
<=> -x + 2 + 3 = 2x
<=> -x + 5 = 2x
<=> -x + 5 - 2x = 0
<=> -3x + 5 = 0
<=> -3x = -5
<=> x = 5/3 ( tmđk )
Vậy nghiệm của phương trình là S = { 5/3 }
* -( x - 2 ) + 3 = 2x ( x < 3 nhé , mình nhầm tí )