a.x+(-35)=18
b.-2x×(-17)=15
c.|x+9|×2=10
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\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)
a. (x + 2)2 + (x + 3)2 - 2(x - 2)(x - 3) = 19
<=> (x2 + 4x + 4) + (x2 + 6x + 9) - (2x + 4)(x - 3) = 19
<=> x2 + 4x + 4 + x2 + 6x + 9 - 2x2 + 6x - 4x + 12 = 19
<=> x2 + x2 - 2x2 + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0
<=> 12x + 6 = 0
<=> 6(2x + 1) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = \(\dfrac{-1}{2}\)
a: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
hay \(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)
a) \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Rightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy x = 2
b) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Rightarrow-3x=-48\)
\(\Leftrightarrow x=16\)
Vậy x = 16
c) \(\dfrac{1}{9}=\dfrac{-2x}{10}\)
\(\Rightarrow-18x=10\)
\(\Leftrightarrow x=-\dfrac{5}{9}\)
Vậy \(x=-\dfrac{5}{9}\)
d) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{x}-5=\dfrac{-9}{x}+2\)
\(\Leftrightarrow\dfrac{3-5x}{x}=\dfrac{-9+2x}{x}\)
\(\Rightarrow3-5x=-9+2x\)
\(\Leftrightarrow7x=12\)
\(\Leftrightarrow x=\dfrac{12}{7}\)
Vậy \(x=\dfrac{12}{7}\)
e) ĐKXĐ: \(x\ne0\)
\(\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Rightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy \(x=\pm4\)
a) Ta có: \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Leftrightarrow-x=\dfrac{6\cdot\left(-8\right)}{3}=-16\)
hay x=16
Vậy: x=16
1) x - 2 = -6
x = - 6 + 2
x = -4
2) -5x-(-3)=13
-5x+3=13
-5x=13-3=10
x=10:(-5)
x=-2
3) 15-(x-7)=21
15-x+7=21
15-x=21-7=14
x=15-14=1
4) 3x+17=2
3x=2-17=-15
x=-15:3=-5
5) 45-(x-9)=-35
45-x+9=-35
45-x=-35-9=-44
x=45-(-44)=89
6) -5+x=15
x=15-(-5)
x=20
7) 2x-(-17)=15
2x+17=15
2x=15-17=-2
x=-2:2=-1
a) x-12=-9+15
x-12=24
x=24+12
x=36
Vậy x=36
b)4x -12=400
4x=412
x=412:4
x=103
Vậy x=103
c)2x -35=15
2x=50
x=50:2
x=15
Vậy x=15
d)3x+17=2
3x=-15
x=(-15):3
x=-5
Vậy x=-5
e)\(\frac{-5}{8}=\frac{x}{16}\)
\(\Rightarrow\left(-5\right).16=8.x\)
\(\Rightarrow8x=-80\)
\(\Rightarrow x=-10\)
f)\(\frac{y}{10}=-\frac{4}{8}\)
\(\Rightarrow8y=\left(-4\right).10\)
\(\Rightarrow8y=-40\)
\(\Rightarrow y=-5\)
a) x - 12 = -9 + 15
=> x - 12 = 6
=> x = 6 + 12 = 18
b) 4x - 12 = 400
=> 4x = 400 + 12 = 412
=> x = 412 : 4 = 103
c) 2x - 35 = 15
=> 2x = 15 + 35 = 50
=> x = 50 : 2 = 25
d) 3x + 17 = 2
=> 3x = 2 - 17
=> 3x = -15
=> x = -15 : 3 = -5
e) \(\frac{-5}{8}=\frac{x}{16}\)
\(=>\frac{-10}{16}=\frac{x}{16}\)
=> x = -10
f) \(\frac{y}{10}=\frac{-4}{8}\)
\(=>\frac{y}{10}=\frac{-1}{2}\)
\(=>\frac{y}{10}=\frac{-5}{10}\)
=> y = -5
Câu tính bn vít lại đề ik, khó hỉu wa
2) a. x2 + 5x = x.(x + 5) âm
=> x.(x + 5) < 0
=> x và x + 5 trái dấu
Mà x < x + 5
=> x < 0; x + 5 > 0
=> x < 0; x > -5
=> x thuộc {-4 ; -3; -2; -1}
b. 3.(2x + 3).(3x - 3)
= 3.(2x + 3).3.(x - 1)
= 9.(2x + 3).(x - 1) âm
=> 9.(2x + 3).(x - 1) < 0
=> (2x + 3).(x - 1) < 0
=> (2x + 3).(2x - 2) < 0
Mà 2x + 3 > 2x - 2
=> 2x + 3 > 0; 2x - 2 < 0
=> 2x > -3; 2x < 2
=>x > -3/2; x < 1
=> x > -2; x < 1
=> x thuộc {-1; 0}
\(a,x+\left(-35\right)=18\)
\(\Rightarrow x=18-\left(-35\right)=18+35=53\)
\(b,-2x.\left(-17\right)=15\)
\(\Rightarrow-2x=-\frac{15}{17}\)
\(\Rightarrow x=\frac{-15}{17}:-2=\frac{-15}{17}.\frac{\left(-1\right)}{2}=\frac{15}{34}\)
\(c,\left|x+9\right|.2=10\)
\(\Rightarrow\left|x+9\right|=10:2=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=\left(-5\right)\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\left(-4\right)\\x=\left(-14\right)\end{cases}}\)
\(a,\)\(x+\left(-35\right)=18\)
\(\Leftrightarrow x=18+35=53\)
\(b,\)\(-2x\left(-17\right)=15\)
\(\Leftrightarrow2x=\frac{15}{17}\)
\(\Leftrightarrow x=\frac{15}{17}\cdot\frac{1}{2}=\frac{15}{34}\)
\(c,\)\(\left|x+49\right|\cdot2=10\)
\(\Leftrightarrow\left|x+49\right|=5\)
\(\Leftrightarrow\hept{\begin{cases}x+49=5\\x+49=-5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-44\\x=-54\end{cases}}\)