B1/Tìm x:
a/ 2x-16=40+x
b/ 16-2x=40-3x
c/ 6(x-1)-5x=15
giải hộ mik nha
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a) ĐK: x ≥ 2
\(\sqrt{3x-6}=3\)
\(\Leftrightarrow3x-6=9\)
<=> 3x = 15
<=> x = 5
Vậy:....
b) ĐK: 5x - 16 ≥ 0
<=> 5x ≥ 16
<=> x ≥ 16/5
\(\sqrt{5x-16}=2\)
<=> 5x - 16 = 4
<=> 5x = 20
<=> x = 4
c) ĐK: \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
bình phương hai vế ta được:
a)điều kiện của x:x≥2
3x-6=9 <=> x=5(nhận)
b)ĐK: x≥16/5
5x-16=4 <=>x=4(nhận)
c) ta có: \(\dfrac{2x-3}{\left(x-2\right)^2-1}\)= \(\dfrac{2x-3}{\left(x-3\right)\left(x-1\right)}\)
ĐKXĐ: x≠3 ;x≠1
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
\(5x-16=40+x\)
\(\Leftrightarrow5x=40+x+16\)
\(\Leftrightarrow5x=x+56\)
\(\Leftrightarrow5x-x=56\)
\(\Leftrightarrow4x=56\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
\(5x-7=-21-2x\)
\(\Leftrightarrow5x-7+21=-2x\)
\(\Leftrightarrow5x+14=-2x\)
\(\Leftrightarrow-2x-5x=14\)
\(\Leftrightarrow-7x=14\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2\)
a.\(x+\dfrac{4}{7}=\dfrac{38}{21}\)
\(x=\dfrac{38}{21}-\dfrac{4}{7}\)
\(x=\dfrac{38}{21}-\dfrac{12}{21}=\dfrac{26}{21}\)
b.\(x-\dfrac{1}{3}=\dfrac{7}{45}:\dfrac{2}{15}\)
\(x-\dfrac{1}{3}=\dfrac{7}{6}\)
\(x=\dfrac{7}{6}+\dfrac{1}{3}\)
\(x=\dfrac{7}{6}+\dfrac{2}{6}=\dfrac{9}{6}\)
a) \(\sqrt{x-3}>2\left(đk:x\ge3\right)\)
\(\Leftrightarrow x-3>4\Leftrightarrow x>7\)
b) \(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-\dfrac{2}{3}\end{matrix}\right.\)
c) \(\sqrt{9\left(5x^2-2x+16\right)}=3x+12\left(đk:x\ge-4\right)\)
\(\Leftrightarrow9\left(5x^2-2x+16\right)=9x^2+72x+144\)
\(\Leftrightarrow36x^2-90x=0\)
\(\Leftrightarrow18x\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
\(a,5x-16=40-1\)
<=>\(5x=40-1+16\)
<=>\(5x=55\)
<=>\(x=11\)
\(b,x-10=-25\)
<=>\(x=\left(-25\right)-10\)
<=>\(x=-35\)
\(c,-12+x=-30\)
<=>\(x=\left(-30\right)+12\)
<=>\(x=-18\)
a) 5x-16=40-1
5x-16=39
5x=39+16
5x=55
x=55:5
x=11
b)x-10=(-25)
x=(-25)+10
x=(-15)
c) -12+x= -30
x= -30-(-12)
x= -30+12
x= -18
d) 2x +12 = -40+ 6
2x+12=(-34)
2x=(-34)-12
2x=(-46)
x=(-46):2
x=(-23)
e)125:(3x-13)=25
3x-13=125:25
3x-13=5
3x=5+13
3x=18
x=18:3
x=6
f)541+(218-x)=735
218-x=735-541
218-x=194
x=218-194
x= 24
g) 3(2x+1)-19=14
3(2x+1)=14+19
3(2x+1)=33
2x+1=33:3
2x+1=11
2x=11-1
2x=10
x=10:2
x=5
h)175-5(x+3)=85
5(x+3)=175-85
5(x+3)=90
x+3=90:5
x+3=18
x=18-3
x=15
i)8x+(-3)=39
8x=39-(-3)
8x=42
x=42:8
x=42/8
L)2x+4x=36+(-6)
6x=30
x=30:6
x=5
a) 7x - 5 = 16 b) 156 - 2x = 82 c) 10x + 65 = 125
=> 7x = 16 + 5 => 2x = 156 - 82 => 10x = 125 - 65
=> 7x = 21 => 2x = 74 => 10x = 60
=> x = 21 : 7 => x = 74 : 2 => x = 60 : 10
=> x = 3 => x = 37 => x = 6
Vậy x = 3 Vậy x = 37 Vậy x = 6
d) 8x + 2x = 25.2 e) 15 + 5x = 40 f) 5x + 2x = 6 - 5
=> 10x = 50 => 5x = 40 - 15 => 7x = 1
=> x = 50 : 10 => 5x = 25 => x = 1 : 7
=> x = 5 => x = 25 : 5 => x = 1/7
Vậy x = 5 => x = 5 Vậy x = 1/7
Vậy x = 5
g) 5x + x = 150 : 2 + 3 h) 6x + 3x = 5 : 5 + 3 i) 5x + 3x = 3 : 3 . 4 + 12
=> 6x = 75 + 3 => 9x = 1 + 3 => 8x = 1 . 4 + 12
=> 6x = 78 => 9x = 4 => 8x = 4 + 12
=> x = 78 : 6 => x = 4 : 9 => 8x = 16
=> x = 13 => x = 4/9 => x = 16 : 8
Vậy x = 13 Vậy x = 4/9 => x = 2
Vậy x = 2
j) 4x + 2x = 68 - 2 : 2 k) 5x + x = 39 - 3 : 3 l) 7x - x = 5 : 5 + 3 . 2 - 7
=> 6x = 68 - 1 => 6x = 39 - 1 => 6x = 1 + 6 - 7
=> 6x = 67 => 6x = 38 => 6x = 7 - 7
=> x = 67 : 6 => x = 38 : 6 => 6x = 0
=> x = 67/6 => x = 19/3 => x = 0
Vậy x = 67/6 Vậy x = 19/3 Vậy x = 0
m) 7x - 2x = 6 : 6 + 44 : 11
=> 5x = 1 + 4
=> 5x = 5
=> x = 5 : 5
=> x = 1
Vậy x = 1
Mỏi tay ~~~~~~~~~~~~~~
a) 2x - 16 = 40 + x
x - 2x = -16 - 40
-x = -56
x = 56
a) 2x - 16 = 40 + x
x - 2x = -16 - 40
-x = -56
x = 56