cau 2 tìm x
a) x - 2/3 = 7/12
b) 1/2x + 3/5 * ( x - 2 ) = 3
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\(2x^3+x^2-4x-12\)
\(=2x^3+5x^2+6x-4x^2-10x-12\)
\(=\left(2x^3+5x^2+6x\right)-\left(4x^2+10x+12\right)\)
\(=x\left(2x^2+5x+6\right)-2\left(2x^2+5x+6\right)\)
\(=\left(x-2\right)\left(2x^2+5x+6\right)\)
\(a,2x^3+x^2-4x-12=\left(2x^3-4x^2\right)+\left(5x^2-10x\right)+\left(6x-12\right)=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)
\(b,x^5-xy^4+x^4y-y^5=x\left(x^4-y^4\right)+y\left(x^4-y^4\right)=\left(x+y\right)\left(x^4-y^4\right)=\left(x+y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)^2\left(x-y\right)\left(x^2+y^2\right)\)
\(c,\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)-9=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]-9=\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9\)
đặt \(x^2+8x+11=y\)
\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9=\left(y-4\right)\left(y+4\right)-9=y^2-16-9=y^2-25=\left(y-5\right)\left(y+5\right)=\left(x^2+8x+11-5\right)\left(x^2+8x+11+5\right)=\left(x^2+8x+6\right)\left(x^2+8x+16\right)=\left(x^2+8x+6\right)\left(x+4\right)^2\)
a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)
\(\Leftrightarrow3x-5\ge2x-24\)
hay \(x\ge-19\)
b: Ta có: \(2\left(5-2x\right)\ge3-x\)
\(\Leftrightarrow10-4x-3+x\ge0\)
\(\Leftrightarrow-3x\ge-7\)
hay \(x\le\dfrac{7}{3}\)
a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\)
<=> \(x=\dfrac{7}{12}-\dfrac{5}{24}=\dfrac{14}{24}-\dfrac{5}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)
b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\)
<=> \(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)
c) bn ghi rõ đề chút
Lời giải:
a. Đề có cả x,y. Bạn xem lại
b.
PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$
$\Leftrightarrow (x-3)(5x-2)=0$
$\Leftrightarrow x-3=0$ hoặc $5x-2=0$
$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$
c.
PT $\Leftrightarrow (7x-2)(x-4)=0$
$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$
$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$
d. Đề thiếu.
a/ x - 7 = 12
=> x = 12 + 7 = 19
b/ 9 + 4(x - 5) = 13
=> 4x - 20 = 4
=> 4x = 24
=> x = 6
c/ (x+2)3 = 64
=> x + 2 = 4
=> x = 2
a) x-7=12
x=12+7
x=19
b) 9+4.(x-5)=13
4.(x-5)=13-9
4.(x-5)=4
(x-5)=4:4
(x-5)=1
x=5+1
x=6
cau cuoi mik ko bt lam nha!
chuc ban hoc tot
1:
a: TH1: x<-3
=>-x-3+10-2x=12
=>-3x+7=12
=>-3x=5
=>x=-5/3(loại)
TH2: -3<=x<5
=>x+3+10-2x=12
=>13-x=12
=>x=1(nhận)
Th3: x>=5
=>x+3+2x-10=12
=>3x=19
=>x=19/3(nhận)
b: =>|2x|+|2x-4|=x+1
TH1: x<0
=>-2x+4-2x=x+1
=>-4x+4-x-1=0
=>-5x=-3
=>x=3/5(loại)
TH2: 0<=x<2
=>2x+4-2x=x+1
=>x=3(loại)
TH3: x>=2
=>2x+2x-4=x+1
=>3x=5
=>x=5/3(loại)
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
a) \(x-\frac{2}{3}=\frac{7}{12}\)
\(x=\frac{7}{12}+\frac{2}{3}\)
\(x=\frac{5}{4}\)
Vậy \(x=\frac{5}{4}\)
b) \(\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)
\(\frac{1}{2}x+\frac{3}{5}x-\frac{3}{5}\times2=3\)
\(\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x=3+\frac{6}{5}\)
\(\frac{11}{10}x=\frac{21}{5}\)
\(x=\frac{42}{11}\)
Vậy \(x=\frac{42}{11}\)
Chúc bạn zui :3
a,x-2/3=7/12
x=7/12+2/3
x=5/4