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1.
$(3^2-2^3)x+3^2.2^2=4^2.3$
$\Leftrightarrow x+36=48$
$\Leftrightarrow x=48-36=12$
2.
$x^5-x^3=0$
$\Leftrightarrow x^3(x^2-1)=0$
$\Leftrightarrow x^3(x-1)(x+1)=0$
$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$
$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.
$(x-1)^2+(-3)^2=5^2(-1)^{100}$
$\Leftrightarrow (x-1)^2+9=25$
$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$
$\Rightarrow x-1=4$ hoặc $x-1=-4$
$\Leftrightarrow x=5$ hoặc $x=-3$
4.
$(2x-1)^2-(2x-1)=0$
$\Leftrightarrow (2x-1)(2x-1-1)=0$
$\Leftrightarrow (2x-1)(2x-2)=0$
$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$
$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$
$\Lef
`@` `\text {Ans}`
`\downarrow`
\((3^2-2^3)x+3^2.2^2=4^2.3\)
`=> x + (3*2)^2 = 48`
`=> x+6^2 = 48`
`=> x + 36 = 48`
`=> x = 48 - 36`
`=> x=12`
Vậy, `x=12`
\(x^5-x^3=0\)
`=> x^3(x^2 - 1)=0`
`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy, `x \in {0; +- 1 }`
\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)
`=> (x-1)^2 + 9 = 25*1`
`=> (x-1)^2 + 9 = 25`
`=> (x-1)^2 = 25 - 9`
`=> (x-1)^2 = 16`
`=> (x-1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy, `x \in {5; -3}`
\((2x-1)^2-(2x-1)=0\)
`=> (2x-1)(2x-1) - (2x-1)=0`
`=> (2x-1)(2x-1-1)=0`
`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
x - 2 | 1 | -1 | 7 | -7 |
x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
1)
a)|17x-5|-|17x+5|=0
17x-5=0 hoặc 17x+5=0
17x=0+5 17x=0-5
17x=5 17x=-5
x=5:17 x=-5:17
x=5/17 x=-5/17
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
a, ta có tổng <0 nên 1 trong 2 số phải có 1 số âm , số còn lại là duong . Mà x-1<x+3 nên x-1 âm và x+3 dưong . Vậy x-1<0 nên x<1;x+3>0nen x>-3.vAY X<1 HOAC X>-3
bạn muốn mình làm cách bth hay lập bảng xét dấu các nhị thức
a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)
b: -2<x<5
B1: Đk: 5x ≥ 0 => x ≥ 0
Vì |x + 1| ≥ 0 => |x + 1| = x + 1
|x + 2| ≥ 0 => |x + 2| = x + 2
|x + 3| ≥ 0 => |x + 3| = x + 3
|x + 4| ≥ 0 => |x + 4| = x + 4
=> |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x
=> x + 1 + x + 2 + x + 3 + x + 4 = 5x
=> 4x + 10 = 5x
=> x = 10
B2: Ta có: |x - 2018| = |2018 - x|
=> A=|x + 2000| + |2018 - x| ≥ |x + 2000 + 2018 - x| = |4018| = 4018
Dấu " = " xảy ra <=> (x + 2000)(x - 2018) ≥ 0
Th1: \(\hept{\begin{cases}x+2000\ge0\\x-2018\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge-2018\\x\le2018\end{cases}}\Rightarrow-2018\le x\le2018\)
Th2: \(\hept{\begin{cases}x+2000\le0\\x-2018\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le-2018\\x\ge2018\end{cases}}\)(vô lý)
Vậy GTNN của A = 4018 khi -2018 ≤ x ≤ 2018
B3:
a, Vì |x + 1| ≥ 0 ; |2y - 4| ≥ 0
=> |x + 1| + |2y - 4| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+1=0\\2y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy...
b, Vì |x - y + 1| ≥ 0 ; (y - 3)2 ≥ 0
=> |x - y + 1| + (y - 3)2 ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y=-1\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3=-1\\y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}\)
Vậy...
c, Vì |x + y| ≥ 0 ; |x - z| ≥ 0 ; |2x - 1| ≥ 0
=> |x + y| + |x - z| + |2x - 1| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\x-z=0\\2x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=z\\x=\frac{1}{2}\end{cases}\Leftrightarrow}}\hept{\begin{cases}\frac{1}{2}+y=0\\x=z=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{-1}{2}\\x=z=\frac{1}{2}\end{cases}}\)
\(\left(x+5\right)\left(x-2\right)\left(x+4\right)\left(x+3\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+5=0\\x-2=0\\x+4=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\\x=-4\\x=3\end{array}\right.\)
giống ảnh đại diện ak
Tìm x :
(x+5)(x−2)(x+4)(x+3)=0
\(\nghiempt{\Rightarrow\begin{cases}x+5=0\\x-2=0\\x+4=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-5\\x=2\\x=-4\\x=-3\end{array}\right.\)