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a) \(\left|7x+3\right|=66\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}7x=63\\7x=-69\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=9\left(N\right)\\x=-\frac{69}{7}\left(L\right)\end{cases}}\)
Vậy...
b) \(\left|5x-2\right|\le0\)
mà \(\left|5x-2\right|\ge0\)
\(\Rightarrow\)\(\left|5x-2\right|=0\)
\(\Leftrightarrow\)\(5x-2=0\)
\(\Leftrightarrow\)\(x=\frac{2}{5}\) (loại)
Vậy...
PT <=> (3x - 1)(6x - 1)(4x - 1)(5x - 1) = 120
. <=> (18x² - 9x + 1)(20x² - 9x + 1) = 120
Đặt a = 19x² - 9x + 1 (Đk a > 0) ta có PT: (a - 1)(a + 1) = 120
<=> a² - 1 = 120
<=> a² = 121
<=> a = 11 (Vì a >0)
Với a = 11 ta có PT: 19x² - 9x - 10 = 0
<=> (10x + 19)(x - 1) = 0
<=> x = 1 (Vì x nguyên)
KL: x = 1
bạn làm sai rồi chỗ hàng 3 ta có: (a-x^2)(a+x^2) mới đúng chứ ko phải (a-1)(a+1)
\(\left|-2x+3\right|-5x=-2\Leftrightarrow\left|2x-3\right|-5x=-2\)
\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\Leftrightarrow2x-3-5x=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x=-\dfrac{1}{2}\left(loai\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x< \dfrac{3}{2}\\-2x+3-5x=-2\end{matrix}\right.\) \(\left\{{}\begin{matrix}x< \dfrac{3}{2}\\-7x=-5=>\left(loai\right)\end{matrix}\right.\)
KL: vô nghiệm:
Bài 1 : a) 3x2 +21x=0
3x(x+7)=0
=> x=0 hoặc x+7=0 =>x=0 hoặc x= -7
b)5x-6x2=0
x(5-6x)=0
=> x=0 hoặc 5-6x=0 => x=0 hoặc x=\(\frac{5}{6}\)
\(3x^2+21x=0\)
\(\Rightarrow3x\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)
\(5x-6x^2=0\)
\(\Rightarrow x\left(5-6x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\5-6x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{6}\end{cases}}}\)
\(\left(2x+3\right)\left(y-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=5\end{cases}}}\)
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
\(a,3-x=8-2x\)
\(\Leftrightarrow x=5\)
Vậy...
\(b,\left|4x+3\right|+5x=5+5x\)
\(\Leftrightarrow\left|4x+3\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}4x+3=5\\4x+3=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0.5\\x=-2\end{cases}}\)
vậy ...
a, 3-x = 8-2x
3-8 = -2x+x
-5 = -x
5 = x
Vậy x=5
b, | 4x+3 | + 5x = 5 + 5x
| 4x+3 | = 5
\(\Rightarrow\orbr{\begin{cases}4x+3=5\\4x+3=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
Vậy............................................
a: =>x(y+1)+y+1=11
=>(x+1)(y+1)=11
=>(x+1;y+1) thuộc {(1;11); (11;1); (-1;-11); (-11;-1)}
=>(x,y) thuộc {(0;10); (10;0); (-2;-12); (-12;-2)}
b: y là số nguyên
=>5x-3 chia hết cho 2x+4
=>10x-6 chia hết cho 2x+4
=>10x+20-26 chia hết cho 2x+4
=>-26 chia hết cho 2x+4
mà x nguyên
nên 2x+4 thuộc {2;-2;26;-26}
=>x thuộc {-1;-3;11;-15}
Vì : \(5x+3⋮2x+1\Rightarrow2\left(5x+3\right)⋮2x+1\Rightarrow10x+6⋮2x+1\)
\(2x+1⋮2x+1\Rightarrow5\left(2x+1\right)⋮2x+1\Rightarrow10x+5⋮2x+1\)
\(\Rightarrow\left(10x+6\right)-\left(10x+5\right)⋮2x+1\)
\(\Rightarrow1⋮2x+1\Rightarrow2x+1\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-1;0\right\}\)