d) 6x³ -8 = 40;
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30 tháng 9 2021
\(100-6x=8^{42}:8^{40}\)
\(\Leftrightarrow100-6x=64\)
\(\Leftrightarrow6x=36\)
hay x=6
BB
2
8 tháng 1 2016
a) x=-72/-8=9
b) x=-54/6=-9
c) x=-40/-4=10
d) x=-66/6=-11
tick nha
8 tháng 1 2016
a)(-8)*x=-72
x=(-72)/(-8)
x=9
b)6*x=-54
x=-54/6
x=-9
c)(-4)*x=(-40)
x=(-40)/(-4)
x=10
d)6x=-66
x=-66/6
x=-11
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
16 tháng 10 2023
\(6x^3-8=40\\ 6x^3=40+8=48\\x^3=\dfrac{48}{6}=8=2^3\\ Vậy:x=2\\ ---\\ 5^{x+1}:5=5^4\\5^{x+1}:5^1=5^4\\ 5^{x+1-1}=5^4\\ 5^x=5^4\\ Vậy:x=4\)
NT
Nguyễn Tuấn Tú
CTVHS
16 tháng 10 2023
\(a,\) \(6x^3-8=40\)
\(6x^3=40+8\)
\(6x^3=48\)
\(x^3=8\)
\(x^3=8\)
\(x^3=2^3\)
\(x=2\)
Vậy \(x=2\)
\(b,\) \(5^{x+1}:5=5^4\)
\(5^{x+1-1}=5^4\)
\(5^x=5^4\)
\(x=4\)
Vậy \(x=4\)
12 tháng 1 2016
a) x= -72 : (-8)
x = 9
b) x = -54 : 6 = -9
c) x = -40 : 9-4) = 10
d) x = -66 : 6 = -11
12 tháng 1 2016
(-8).x= -72
x=(-72):(-8)
x=9
6x=-54
x=(-54):6
x=-9
(-4).x=-40
x=40:(-4)
x= -10
6x=-66
x=(-66):6
x= -11
tick mk nha p!
PT
10 tháng 9 2018
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
PT
10 tháng 9 2018
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
26 tháng 10 2017
\(A=16x^2+8x+3\\ A=16x^2+8x+1+2\\ A=\left(16x^2+8x+1\right)+2\\ A=\left(4x+1\right)^2+2\\ Do\left(4x+1\right)^2\ge0\forall x\\ \Rightarrow A=\left(4x+1\right)^2+2\ge2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(4x+1\right)^2=0\\ \Leftrightarrow4x+1=0\\ \Leftrightarrow4x=-1\\ \Leftrightarrow x=-\dfrac{1}{4}\\ \text{Vậy }A_{\left(Min\right)}=2\text{ khi }x=-\dfrac{1}{4}\\ \)
\(B=y^2-5y+8\\ B=y^2-5y+\dfrac{25}{4}+\dfrac{7}{4}\\ B=\left(y^2-5y+\dfrac{25}{4}\right)+\dfrac{7}{4}\\ B=\left[y^2-2\cdot y\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{7}{4}\\ B=\text{ }\left(y-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\\ Do\text{ }\left(y-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow B=\left(y-\dfrac{5}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(y-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{5}{2}=0\\ \Leftrightarrow y=\dfrac{5}{2}\\ \text{Vậy }B_{\left(Min\right)}=\dfrac{7}{4}\text{ }khi\text{ }y=\dfrac{5}{2}\)
\(C=2x^2-2x+2\\ C=2x^2-2x+\dfrac{1}{2}+\dfrac{3}{2}\\ C=\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{3}{2}\\ C=2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{3}{2}\\ C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{1}{2}=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{3}{2}\text{ }khi\text{ }x=\dfrac{1}{2}\)
\(D=9x^2-6x+25y^2+10y+4\\ D=9x^2-6x+25y^2+10y+1+1+2\\ D=\left(9x^2-6x+1\right)+\left(25y^2+10y+1\right)+2\\ D=\left[\left(3x\right)^2-2\cdot3x\cdot1+1^2\right]+\left[\left(5y\right)^2+2\cdot5y\cdot1+1^2\right]+2\\ D=\left(3x-1\right)^2+\left(5y+1\right)^2+2\\ Do\text{ }\left(3x-1\right)^2\ge0\forall x\\ \left(5y+1\right)^2\ge0\forall y\\ \Rightarrow\left(3x-1\right)^2+\left(5y+1\right)^2\ge0\forall x;y\\ \Rightarrow D=\left(3x-1\right)^2+\left(5y+1\right)^2+2\ge2\forall x;y\\ \text{Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(3x-1\right)^2=0\\\left(5y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\5y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=1\\5y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{5}\end{matrix}\right.\\ \text{Vậy }D_{\left(Min\right)}=2\text{ khi }x=\dfrac{1}{3};y=-\dfrac{1}{5}\)
26 tháng 10 2017
Câu 2
\(M=x^2+6x+1\\ M=x^2+6x+9-8\\ M=\left(x^2+6x+9\right)-8\\ M=\left(x+3\right)^2-8\\ Do\text{ }\left(x+3\right)^2\ge0\forall x\\ M=\left(x+3\right)^2-8\ge-8\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\\ \text{Vậy }M_{\left(Min\right)}=-8\text{ khi }x=-3\)
\(N=10y-5y^2-3\\ N=10y-5y^2-5+2\\ N=-\left(5y^2-10y+5\right)+2\\ N=-5\left(y^2-2y+1\right)+2\\ N=-5\left(y-1\right)^2+2\\ Do\left(y-1\right)^2\ge0\forall x\\ \Rightarrow-\left(y-1\right)^2\le0\forall x\\ \Rightarrow-5\left(y-1\right)^2\le0\forall x\\ \Rightarrow N=-5\left(y-1\right)^2+2\le2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(y-1\right)^2=0\\ \Leftrightarrow y-1=0\\ \Leftrightarrow y=1\\ \text{Vậy }N_{\left(Max\right)}=2\text{ khi }y=1\)
KT
4
HB
16 tháng 10 2020
\(a.5^x=125\)
\(\Leftrightarrow5^x=5^3\)
\(\Leftrightarrow x=3\)
\(b.6x^3-8=40\)
\(\Leftrightarrow6x^3=40+8=48\)
\(\Leftrightarrow x^3=\frac{48}{6}=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
\(c.\left(x+1\right)^3=64\)
\(\Leftrightarrow\left(x+1\right)^3=4^3\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=4-1=3\)
\(d.3^x\div3^2=243\)
\(\Leftrightarrow3^{x-2}=3^5\)
\(\Leftrightarrow x-2=5\)
\(\Leftrightarrow x=5+2=7\)
20 tháng 2 2021
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
\(6x^3-8=48\\ =>6x^3=40+8\\ =>6x^3=48\\ =>x^3=48:6\\ =>x^3=8\\ =>x^3=2^3\\ =>x=2\)
\(6x^3-8=40\)
\(6x^3=48\)
\(x^3=8\)
\(x^3=2^3\)
\(x=2\)