\(x\) E BC (16 ,2\(x\) ) ; \(x\) < 50
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Lời giải:
a.
$23+x=15$
$x=15-23=-8$
b.
$x-18=-23$
$x=-23+18=-5$
c.
$x+(-4)=16$
$x=16-(-4)=16+4=20$
d.
$25-x=-16$
$x=25-(-16)=25+16=41$
e.
$(-35)x=-210$
$x=(-210):(-35)=6$
f.
$(-2)(15-x)=60$
$15-x=60:(-2)=-30$
$x=15-(-30)=15+30=45$
x^20-x=0
x(x^19-1)=0
x= 0
hoặc x ^ 19 =1
x = 0 hoặc x= 1
\(a,x^2-y^2-2x+2y=\left(x^2-y^2\right)-\left(2x-2y\right)=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right).\) \(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\)
\(c,3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3.\left(\left(a-b\right)^2-\left(2c\right)^2\right)\)
\(=3\left(a-b-2c\right).\left(a-b+2c\right)\)
\(d,x^2-25+y^2-2xy=\left(x^2-2xy+y^2\right)-5^2=\left(x-y\right)^2-5^2\)
\(=\left(x-y+5\right)\left(x-y-5\right)\)
\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
\(f,x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\(h,x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x-1\right)\left(x^2-16\right)=\)
\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)
c) \(\left(3x+2\right)^3=-27\)
\(\Rightarrow\left(3x+2\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+2=-3\)
\(\Rightarrow3x=\left(-3\right)-2\)
\(\Rightarrow3x=-5\)
\(\Rightarrow x=\left(-5\right):3\)
\(\Rightarrow x=-\frac{5}{3}\)
Vậy \(x=-\frac{5}{3}.\)
Chúc bạn học tốt!
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
x^3 = 27
x^3 = 3^3
=> x = 3
(2x-1)^3 = 8
(2x-1)^3 = 2^3
2x-1 = 2
2x = 2+1
2x = 3
x = 3:2
=> ko có x phù hợp.
(x-2)^2 = 16
(x-2)^2 = 4^2
x-2 = 4
x = 4+2
x = 6
Chúc bn học tốt!
x^3 = 27
x^3 = 3^3
Vậy x = 3
Đề bài 2 hình như sai bạn ạ
( x - 2 )^2 = 16
( x- 2 )^2 = 4^2
x - 2 = 4
x = 4 + 2
x = 6
Vậy x = 6