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Bài 1. Giải các phương trình sau
a) \(5\left(x-2\right)=3\left(x+1\right)\)
\(\Leftrightarrow5x-10=3x+3\)
\(\Leftrightarrow5x-3x=10+3\)
\(\Leftrightarrow2x=13\)
\(\Leftrightarrow x=\dfrac{13}{2}\)
Vậy \(S=\left\{\dfrac{13}{2}\right\}\)
b) \(\dfrac{2x}{x+1}+\dfrac{3}{x-2}=2\left(1\right)\)
Điều kiện: \(x+1\ne0\Leftrightarrow x\ne-1\) và \(x-2\ne0\Leftrightarrow x\ne2\)
\(\left(1\right)\Leftrightarrow\dfrac{2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{2\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2x\left(x-2\right)+3\left(x+1\right)=2\left(x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2x^2-4x+3x+3=2x^2-4x+2x-4\)
\(\Leftrightarrow2x^2-4x+3x-2x^2+4x-2x=-3-4\)
\(\Leftrightarrow x=-7\left(N\right)\)
Vậy \(S=\left\{-7\right\}\)
c) \(|2x+7|=3\)
\(\Leftrightarrow2x+7=3\) hoặc \(2x+7=-3\)
.. \(2x+7=3\Leftrightarrow2x=-4\Leftrightarrow x=-2\)
.. \(2x+7=-3\Leftrightarrow2x=-10\Leftrightarrow x=-5\)
Vậy \(S=\left\{-2;-5\right\}\)
Bài 2 bạn ghi rõ đề lại nha r mik giải lun cho
Bài 2. Giải các bất phương trình sau:
a) \(\left(x+2\right)^2< \left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow x^2+4x+4< x^2-1\)
\(\Leftrightarrow x^2+4x-x^2< -4-1\)
\(\Leftrightarrow4x< -5\)
\(\Leftrightarrow x>-\dfrac{5}{4}\)
Vậy \(S=\left\{x/x< -\dfrac{5}{4}\right\}\)
Câu b mik tính ko ra nhá sorry!!!!!!!!!!
Ta khai triển VT trước
\(VT=\frac{1-b-c+bc}{b+c}+\frac{1-c-a+ca}{c+a}+\frac{1-a-b+ab}{a+b}=\frac{\left(1-b\right)-c\left(1-b\right)}{1-a}+\frac{\left(1-c\right)-a\left(1-c\right)}{1-b}+\frac{\left(1-a\right)-b\left(1-a\right)}{1-c}=\frac{\left(1-c\right)\left(1-b\right)}{1-a}+\frac{\left(1-c\right)\left(1-a\right)}{1-b}+\frac{\left(1-a\right)\left(1-b\right)}{1-c}\)Với a,b,c luôn dương vào a+b+c=1 nên a,b,c<1\(\Rightarrow\)1-a,1-b,1-c>0
Áp dụng Cosi có \(\frac{\left(1-c\right)\left(1-b\right)}{1-a}+\frac{\left(1-c\right)\left(1-a\right)}{1-b}\ge2\left(1-c\right)\left(1\right)\).Tương tự: \(\frac{\left(1-c\right)\left(1-a\right)}{1-b}+\frac{\left(1-a\right)\left(1-b\right)}{1-c}\ge2\left(1-a\right)\left(2\right)\)
\(\frac{\left(1-c\right)\left(1-b\right)}{1-a}+\frac{\left(1-a\right)\left(1-b\right)}{1-c}\ge2\left(1-b\right)\left(3\right)\)
Cộng (1),(2) và (3) có \(2VT\ge2\left(3-a-b-c\right)\Leftrightarrow VT\ge3-1=2\)
1)
$x^3+9x^2+23x+15=(x^3+x^2)+(8x^2+8x)+(15x+15)$
$=x^2(x+1)+8x(x+1)+15(x+1)$
$=(x+1)(x^2+8x+15)$
$=(x+1)[(x^2+3x)+(5x+15)]$
$=(x+1)[x(x+3)+5(x+3)]=(x+1)(x+3)(x+5)$
5)
$x^4+5x^2+9=(x^4+6x^2+9)-x^2$
$=(x^2+3)^2-x^2=(x^2+3-x)(x^2+3+x)$
3)
$(3x-2)^2(6x-5)(6x-3)-5$
$=(9x^2-12x+4)(36x^2-48x+15)-5$
$=(9x^2-12x+4)[4(9x^2-12x)+15]-5$
$=(a+4)(4a+15)-5$ (đặt $9x^2-12x=a$)
$=4a^2+31a+55$
$=4a^2+20a+11a+55$
$=4a(a+5)+11(a+5)=(4a+11)(a+5)=(36x^2-48x+11)(9x^2-12x+5)$
$=
\(x^3-2x^2-x+2\)
\(=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)