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\(a,2.\left(x+15\right)-5=35\)
\(\Leftrightarrow2.\left(x+15\right)=35+5\)
\(\Leftrightarrow2.\left(x+15\right)=40\)
\(\Leftrightarrow x+15=40:2\)
\(\Leftrightarrow x+15=20\)
\(\Leftrightarrow x=20-15\)
\(\Leftrightarrow x=5\) \(\text{Vậy }x=5\)
\(b,9x-33=3^{2015}:3^{2014}\)
\(\Leftrightarrow9x-33=3^{2015-2014}\)
\(\Leftrightarrow9x-33=3\)
\(\Leftrightarrow9x=3+33\)
\(\Leftrightarrow9x=36\)
\(\Leftrightarrow x=36:9\)
\(\Leftrightarrow x=4\) \(\text{Vậy }x=4\)
\(c,3.\left(2x-4\right)^3=24\)
\(\Leftrightarrow\left(2x-4\right)^3=24:3\)
\(\Leftrightarrow\left(2x-4\right)^3=8\)
\(\Leftrightarrow\left(2x-4\right)^3=2^3\)
\(\Leftrightarrow2x-4=2\)
\(\Leftrightarrow2x=2+4\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=6:2\)
\(\Leftrightarrow x=3\) \(\text{Vậy }x=3\)
\(d,\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=21+140\)
\(\Leftrightarrow x=161\) \(\text{Vậy }x=161\)
\(e,2.3^x-5^3:5^5=29\)
\(\Leftrightarrow2.3^x-5^3:5^3:5^2=29\)
\(\Leftrightarrow2.3^x-1:5^2=29\)
\(\Leftrightarrow2.3^x-\frac{1}{25}=29\)
\(\Leftrightarrow2.3^x=29+\frac{1}{25}\)
\(\Leftrightarrow2.3^x=\frac{726}{25}\)
\(\Leftrightarrow3^x=\frac{726}{25}:2\)
\(\Leftrightarrow3^x=\frac{726}{25}:2\)
\(\Leftrightarrow3^x=\frac{363}{25}\)
\(\Leftrightarrow3^x=3^{2,435369636...}\)
\(\Leftrightarrow x=2,435369636...\)
\(\text{Vậy }x=2,435369636\left(\text{bạn thử kiểm tra lại đề nha}\right)\)
\(f,12:x\text{ và }x\le5\)
\(\text{Do }x\le5\)
\(\Leftrightarrow12:x\le2,4\left(\text{lấy 12 chia cho cả 2 vế}\right)\)
\(\text{Câu này bạn phải cho điều kiện }x\in N,\text{nếu không x có nhiều giá trị lắm }\)
\(\text{Nếu }x\in N\Rightarrow12:x\in N\)
\(\Leftrightarrow12:x\le2,4\left(\text{lấy 12 chia cho cả 2 vế}\right)\)
\(\Rightarrow12:x\in\left\{0;1;2\right\}\)
\(\begin{matrix}12:x&0&1&2\\x&x\in\varnothing&12&6\\\text{Đánh giá}&Loại&Chọn&Chọn\end{matrix}\)
\(\text{Vậy }x\in\left\{6;12\right\}\)
\(g,3x-2x+7x=16\)
\(\Leftrightarrow x.\left(3-2+7\right)=16\)
\(\Leftrightarrow x.8=16\)
\(\Leftrightarrow x=16:8\)
\(\Leftrightarrow x=2\) \(\text{Vậy }x=2\)
\(h,5^{10}.\left(2x-1\right)=5^{12}\)
\(\Leftrightarrow2x-1=5^{12}:5^{10}\)
\(\Leftrightarrow2x-1=5^2\)
\(\Leftrightarrow2x-1=25\)
\(\Leftrightarrow2x=25+1\)
\(\Leftrightarrow2x=26\)
\(\Leftrightarrow x=26:2\)
\(\Leftrightarrow x=13\) \(\text{Vậy }x=13\)
\(i,5⋮\left(x+1\right)\)
\(\Rightarrow x+1\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)
x+1 | 1 | -1 | 5 | -5 |
x | 0 | -2 | 4 | -6 |
Đánh giá | Chọn | Chọn | Chọn | Chọn |
\(\text{Vậy }x\in\left\{-6;-2;0;4\right\}\)
a) 5x * 5 = 54
5x+1 = 54
=> x = 3
c) (x-1)2=25
(x-1)2=52
x - 1 = 5
x = 6
a, 5x . 5 = 54
=> 5x + 1 = 54
=> x + 1 = 4
=> x = 3
vậy_
câu b đề k rõ
c, (x - 1)2 = 25
=> (x - 1)2 = (+ 5)2
=> x - 1 = + 5
=> x = 6 hoặc x = -4
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
a) (x - 2)(x + 4) = 0
=> x - 2 = 0 hoặc x + 4 = 0
=> x = 2 hoặc x = -4
b) Đề lỗi
c) 72 - 3|x + 1| = 9
=> 3|x + 1| = 72 - 9
=> 3|x + 1| = 63
=> |x + 1| = 63 : 3
=> |x + 1| = 21
=> x + 1 = 21 hoặc x + 1 = -21
=> x = 20 hoặc x = -22
d) 7(x - 3) - 5(3 - x) = 11x - 5
=> 7x - 21 - 15 + 5x = 11x - 5
=> 7x - 21 - 15 + 5x - 11x + 5 = 0
=> x - 31 = 0
=> x = 31
2) \(\left(x-2\right).\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
vậy \(x=2\) hoặc \(x=-4\)
3) \(\left(x-2\right).\left(x+15\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-15\end{cases}}}\)
vậy \(x=2\) hoặc \(x=-15\)
4) \(\left(7-x\right).\left(x+19\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-x=0\\x+19=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-19\end{cases}}}\)
vậy \(x=7\) hoặc \(x=-19\)
8) \(2x^2-3=29\)
\(2x^2=29+3\)
\(2x^2=32\)
\(x^2=32\div2\)
\(x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x^2=4^2\\x^2=\left(-4\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(x=4\) hoặc \(x=-4\)
7)Ta có : x - 3 = 0 x - 5 = 0
=> x = 0 + 3 => x = 0 +5
=> x = 3 => x = 5
Ta lập bảng xét dấu :
x | 3 | 5 | |||
x-3 | - | 0 | + | + | |
x-5 | - | - | 0 | + | |
(x-3).(x-5) | + | 0 | - | 0 | + |
Vậy để (x-3).(x-5) < 0 => 3<x<5 => x = 4
6) | x | <3
=>x thuộc cộng trừ 1 , cộng trừ 2
a) \(12x-60+21-7x-5=0\)
\(\Rightarrow5x-44=0\)
\(\Rightarrow5x=44\)
\(\Rightarrow x=\frac{44}{5}\)
Vậy: \(x=\frac{44}{5}\)
b) \(2x^2-3=29\)
\(\Rightarrow2x^2=3+29=32\)
\(\Rightarrow x^2=\frac{32}{2}=16\)
\(\Rightarrow x=\sqrt{16}=4\)
Vậy: \(x=4\)