3x+29=x+(-9)
3x – 47.2 + x = - 29 + (-9)
3x – 47.2 + x = - 29 + (-9)
3x – 47.2 + x = -38
3x – 94 + x = -38
3x – 94 = -38 - x
3x + x = -38 + 94
4x = 56
x = 56 : 4
x = 14
Vậy x = 14
\(3x-47.2+x=-29+\left(-9\right)\\ \Rightarrow4x-94=-38\\ \Rightarrow4x=-38+94\\ \Rightarrow4x=56\\ \Rightarrow x=56:4\\ \Rightarrow x=14\)
Tìm số nguyên x, biết:
3x - 47.2 + x = -29 + -9
Tìm x, biết a. 0,5x -2/3x=7/12 b. -8/17+5/17<x/17<-6/17+9/17 c. [x-5/12].9/29=-6/29
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
b)\(\frac{-8}{17}+\frac{5}{17}< \frac{x}{17}< \frac{-6}{17}+\frac{9}{17}\)\(\left(x\in Z\right)\)
\(\Rightarrow\frac{-3}{17}< \frac{x}{17}< \frac{3}{17}\)
\(\Rightarrow-3< x< 3\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
Vậy \(x\in\left\{-2;-1;0;1;2\right\}\)
Tìm x ∈ Z, biết:
a) x + 5 = -2 + 11
b) -3x = -5 + 29
c) | x | - 9 = -2 + 17
d) | x – 9 | = -2 + 17
a) x = 4
b) x = -8
c) | x | - 9 = -2 + 17
| x | = 15 + 9
| x | = 24
x = 24 hoặc x = -24
d) |x – 9| = -2 + 17
|x – 9| = 15
x – 9 = 15 hoặc x – 9 = -15
x = 24 hoặc x = -6
Đề: a/ (3x-x)^2 (3x+1) (3x+1)=29
b/(4x-1)+(9-4x) (3+4x)=-8
a, \(\left(3x-x\right)^2\left(3x+1\right)\left(3x+1\right)=29\)
<=> \(4x^2\left(3x+1\right)^2=29\)
<=> \(4x^2;\left(3x+1\right)^2\inƯ\left(29\right)=\left\{\pm1;\pm29\right\}\)
4x^2 | 1 | -1 | 29 | -29 |
(3x+1)^2 | 29 | -29 | 1 | -1 |
x | 1/2 | ktm | \(\sqrt{\frac{29}{4}}\) | ktm |
x | \(\frac{\sqrt{29}-1}{3}\) | ktm | 0 | ktm |
b, Tương tự
b) ( 4x - 1 ) + ( 9 - 4x )( 3 + 4x ) = -8
<=> ( 4x - 1 ) + ( 27 + 24x - 16x2 ) = -8
<=> 4x - 1 + 27 + 24x - 16x2 = -8
<=> -16x2 + 28x + 26 = -8
<=> -16x2 + 28x + 26 + 8 = 0
<=> -16x2 + 28x + 34 = 0
<=> -2( 8x2 - 14x - 17 ) = 0
=> 8x2 - 14x - 17 = 0
\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=\left(\frac{-14}{2}\right)^2-\left(-17\right)\cdot8=185\)
\(\Delta'>0\)nên phương trình đã cho có hai nghiệm phân biệt :
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-\left(-7\right)+\sqrt{185}}{8}=\frac{7+\sqrt{185}}{8}\)
\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-\left(-7\right)-\sqrt{185}}{8}=\frac{7-\sqrt{185}}{8}\)
Lớp 7 mà nghiệm xấu nhỉ ?
A = a^3 + 1 + 3a + 3a^2 với a = 9
B = x^3 + 3x^2 + 3x + 1 với x = 19
C = a^3 + 3a^2 + 3a + 6 với a = 29
a: \(A=\left(a+1\right)^3=10^3=1000\)
b: \(B=\left(x+1\right)^3=20^3=8000\)
c: \(C=a^3+3a^2+3a+1+5\)
\(=30^3+5=27005\)
Cho A = (1 x 2 x 3x....x 48 x 49) : (9 x 19 x 29 x 39 x 49) . Tính tổng 10 chữ số tận cùng của A
Tìm x:
a) 3x + 17 = 2
b) 2x + 11 = 3 (x - 9)
2x2 - 3 = 29
b) 2x+11=3(x-9)
=> 2x+11= 3x-27
=> 2x=3x-38
=> 2x+38=3x
=> 38=3x-2x
=> 38=x
c) 2x2-3=29
=> 2x2=29+3
=> 2x2=32
Vì ko có số bào mà bình phương của nó bằng 32
=> x ko tồn tại
7-(2x-1/3)^2=3
(2x+1/3)^2-3/8=1/8
12:[29-(x-2/3)^2]=3
(3x-1/2)^3+8/3=29/9-14/27
2(2x-1/3)^2+4/3=5/6+13/18
Tìm X