Tìm x biết
1)(x-0)^6-(x-2)^8
2)2^x+2^x+1+2^x+2+2^x+3=480
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TT
1
HP
14 tháng 10 2021
1. x(x + 1) - x2 + 1 = 0
<=> x(x + 1) - (x2 - 1) = 0
<=> x(x + 1) - (x + 1)(x - 1) = 0
<=> (x - x + 1)(x + 1) = 0
<=> x + 1 = 0\
<=> x = -1
2. 4x(x - 2) - 6 + 3x = 0
<=> 4x(x - 2) - (3x - 6) = 0
<=> 4x(x - 2) - 3(x - 2) = 0
<=> (4x - 3)(x - 2) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)
3. x(x + 2) - 3(x + 2) = 0
<=> (x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
14 tháng 11 2023
Lời giải:
1. $(x+2)-2=0$
$x+2=2$
$x=0$
2.
$(x+3)+1=7$
$x+3=7-1=6$
$x=6-3=3$
3.
$(3x-4)+4=12$
$3x-4+4=12$
$3x=12$
$x=12:3=4$
4.
$(5x+4)-1=13$
$5x+4=13+1=14$
$5x=14-4=10$
$x=10:5=2$
5.
$(4x-8)-3=5$
$4x-8=5+3=8$
$4x=8+8=16$
$x=16:4=4$
6.
$3+(x-5)=7$
$x-5=7-3=4$
$x=4+5=9$
7.
$8-(2x-4)=2$
$2x-4=8-2=6$
$2x=6+4=10$
$x=10:2=5$
8.
$7+(5x+2)=14$
$5x+2=14-7=7$
$5x=7-2=5$
$x=5:5=1$
9.
$5-(3x-11)=1$
$3x-11=5-1=4$
$3x=11+4=15$
$x=15:3=5$
10.
$16-(8x+2)=6$
$8x+2=16-6=10$
$8x=10-2=8$
$x=8:8=1$
NH
25 tháng 12 2022
1)
\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(3x-6+4x-4=25\)
\(7x-10=25\\ 7x=35\\ x=5\)
2)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\left(x-2\right)\left(4x-2\right)=0\)
\(=>\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
3)
\(\left(x-2\right)^2=4\left(x-1\right)^2\)
\(x^2-4x+4=4\left(x^2-2x+1\right)\)
\(x^2-4x+4=4x^2-8x+4\)
\(x^2-4x+4-4x^2+8x-4=0\)
\(-3x^2+4x=0\)
\(x\left(-3x+4\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\-3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
14 tháng 10 2021
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
30 tháng 9 2021
1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2
3)
(2x+1)2=52
2x+1=5
2x=4
x=2
30 tháng 9 2021
\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)
\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
26 tháng 11 2021
\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
26 tháng 11 2021
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
HP
0
câu 1 sai đề :>>>
1) \(\left(x-2\right)^6=\left(x-2\right)^8\)
\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)
\(\Leftrightarrow\left(x-2\right)^6.\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3,x=1\end{cases}}\)
2) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Leftrightarrow2^x.\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow2^x.15=480\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Leftrightarrow x=5\)