tỉm x:
a) x-36=-15
b) 5.(x-37)=-80
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a)x.(5-2x)-2x.(1-x)=15
x [ 5 - 2x -2.(1-x) ] = 15
x ( 5 - 2x -2 + 2x ) =15
x . 3 =15
x = 5
b)(3x+2)2+(1+3x).(1-3x)=2
9x2+12x+4+1-9x2=2
12x + 5 = 2
12x = -3
x = -1/4
Bài 1:
a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Bài 2:
a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)
\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)
\(a,x-36:18=12-15\\ \Rightarrow x-2=-3\\ \Rightarrow x=-1\\ b,92-\left(17+x\right)=72\\ \Rightarrow17+x=20\\ \Rightarrow x=3\\ c,720:\left[41-\left(2x+5\right)\right]=40\\ \Rightarrow41-\left(2x+5\right)=18\\ \Rightarrow2x+5=23\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ d,\left(x+2\right)^3-23=41\\ \Rightarrow\left(x+2\right)^3=64\\ \Rightarrow\left(x+2\right)^3=4^3\\ \Rightarrow x+2=4\\ \Rightarrow x=2\)
a. 2x + 70 = 74
<=> 2x = 4
<=> x = 2
b. 120 - \(\dfrac{4x}{2}\) = 80
<=> 120 - 2x = 80
<=> 120 - 80 = 2x
<=> 2x = 40
<=> x = 20
c. (3x + 5)2 = 400
<=> \(|3x+5|=\sqrt{400}\)
<=> \(|3x+5|=20\)
<=> \(\left[{}\begin{matrix}3x+5=20\\3x+5=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-25}{3}\end{matrix}\right.\)
Trong các số có 3 chữ số gióng nhau chỉ có 111 là chia hết cho 37 => aaa(ngang)=111 => a=1
Ta có: 111:(x:1)=37
<=> x:1= 111:37
<=> x:1= 3
<=> x= 3x1
<=> x=3
a.
$x^4-25x^3=0$
$\Leftrightarrow x^3(x-25)=0$
\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)
b.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
c.
$x^3-4x^2-9x+36=0$
$\Leftrightarrow x^2(x-4)-9(x-4)=0$
$\Leftrightarrow (x-4)(x^2-9)=0$
$\Leftrightarrow (x-4)(x-3)(x+3)=0$
\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)
d. ĐK: $x\neq 0$
$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$
$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$
$\Leftrightarrow -2(-x^2+3x-4)=0$
$\Leftrightarrow x^2-3x+4=0$
$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)
Vậy pt vô nghiệm.
\(a,x.4,25=7\\ \Rightarrow x=7:4,25\\ \Rightarrow x=\dfrac{28}{17}\\ b,116:x=14,5\\ \Rightarrow x=116:14,5\\ \Rightarrow x=8\\ c,15.x=8\\ \Rightarrow x=\dfrac{8}{15}\\ c,729:x=36\\ \Rightarrow x=729:36\\ \Rightarrow x=20,25\)
\(a,25x^2-1=15\)\(< =>x^2=\dfrac{16}{25}< =>x=\pm\dfrac{4}{5}\)
\(b,\left(x-4\right)^2-\left(5x+2\right)^2=0\)\(< =>\left(-4x-6\right)\left(6x-2\right)=0\)
\(< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(c,\left(x-1\right)\left(x-9\right)=0< =>\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Lời giải:
a. $21-2(x+6)=15$
$2(x+6)=21-15=6$
$x+6=6:2=3$
$x=3-6=-3$
b.
$3^{x+1}=27=3^3$
$\Rightarrow x+1=3$
$\Rightarrow x=2$
c.
$14-x=36$
$x=14-36=-22$
d.
$-3x+25=-2$
$-3x=-2-25=-27$
$x=(-27):(-3)=9$
e.
$37-2(x+6)=29$
$2(x+6)=37-29=8$
$x+6=8:2=4$
$x=4-6=-2$
h.
$2^{x-1}=16=2^4$
$x-1=4$
$x=4+1=5$
a. x - 36 = -15
=> x = -15 + 36
=> x = 21
b. 5.(x - 37) = -80
=> x - 37 = -80 : 5
=> x - 37 = -16
=> x = -16 + 37
=> x = 21
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