Tìm x [-300]:20+5.[3x-1]=25
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Mình làm câu khó thôi nhé.
2x chia hết cho 3
=>(2x+x-x) chia hết cho 3
=>(3x-x) chia hết cho 3
3x chia hết cho 3=>x chia hết cho 3
=>x thuộc B(3)={0;3;6;...}
Vậy x thuộc {0;3;6;...ư}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-300\right):20+5.\left(3x-1\right)=25\)
\(\Rightarrow-15+15x-5=25\)
\(\Rightarrow-20x+15x=25\)
\(\Rightarrow15x=45\)
\(\Rightarrow x=3\)
( - 300 ) : 20 + 5( 3x -1 ) = 25
- 15 + 5( 3x -1 ) = 25
-15 + 15x -5 = 25
- 20 + 15x = 25
15x - 20 = 25
15x - 20 + 20 = 25 + 20
15x = 25 + 20
15x = 45
15x : 15 = 45 : 15
x = 45 : 15
x = 3
vậy x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
a: \(35\cdot16+35\cdot28-44\cdot15\)
\(=35\left(16+28\right)-44\cdot15\)
\(=44\left(35-15\right)\)
\(=44\cdot20=880\)
b: \(240-2\left(3\cdot5^2-20:2^2\right)\)
\(=240-2\left(3\cdot25-20:4\right)\)
\(=240-150+10=10+90=100\)
2:
b: \(\left(8-3x\right)^4-1=15\)
=>\(\left(3x-8\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-8=2\\3x-8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=6\end{matrix}\right.\)
=>x=10/3 hoặc x=2
c: \(218-5\left(x-8\right)=2^5:2^2\)
=>\(218-5\left(x-8\right)=2^3=8\)
=>5(x-8)=210
=>x-8=42
=>x=50
d: \(\left(5-3x\right)^4-1=15\)
=>\(\left(3x-5\right)^4-1=15\)
=>\(\left(3x-5\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-5=-4\\3x-5=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)
\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)
\(\Leftrightarrow-297-99x=0\)
\(\Leftrightarrow x=3\)
Vậy \(n_0\) của PT là: x=3
b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)
\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)
\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)
\(\Leftrightarrow-64-36x=250-30x\)
\(\Leftrightarrow-6x=314\)
\(\Leftrightarrow x=-\frac{157}{3}\)
Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)
c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)
\(\Leftrightarrow-3x=4x-\frac{23}{5}\)
\(\Leftrightarrow7x=\frac{23}{5}\)
\(\Leftrightarrow x=\frac{23}{35}\)
Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)
d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow x=-\frac{5}{12}\)
Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x2+x-6)(x2+9x+14) = 300
<=> (x-2)(x+3)(x+2)(x+7) - 300 = 0
<=> [(x-2)(x+7)][(x+2)(x+3)] - 300 = 0
<=> (x2-5x-14)(x2+5x+6) - 300 = 0
Đặt x2 + 5x - 14 = a
<=> a(a+20) - 300 = 0
<=> a2 + 20a - 300 = 0
<=> a2 + 20a + 100 - 400 = 0
<=> (a+10)2 - 202 = 0
<=> (a-10)(a+30) = 0
<=> \(\left[{}\begin{matrix}a=10\\a=-30\end{matrix}\right.\)
Với a = 10, ta có:
x2 + 5x - 14 = 10
=> x2 + 5x - 24 = 0
=> (x-3)(x+8) = 0
=> \(\left[{}\begin{matrix}x=3\\x=-8\end{matrix}\right.\)
Với a = -30, ta có:
x2 + 5x - 14 = -30
=> x2 + 5x + 16 = 0 (vn)
Vậy nghiệm pt x = 3; x = -8
b) (2x-5)(3x+1) = 4x2 - 25
<=> (2x-5)(3x+1) = (2x-5)(2x+5)
<=> (2x-5)(3x+1-2x-5) = 0
<=> (2x-5)(x-4) = 0
<=> \(\left[{}\begin{matrix}2x-5=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=4\end{matrix}\right.\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với x ≥ 0 , x ≠ 25 Ta có: A = B . x − 4
⇔ x + 2 x − 5 = 1 x − 5 . x − 4 ⇔ x + 2 = x − 4 ( * )
Nếu x ≥ 4 , x ≠ 25 thì (*) trở thành : x + 2 = x − 4
⇔ x − x − 6 = 0 ⇔ x − 3 x + 2 = 0
Do x + 2 > 0 nên x = 3 ⇔ x = 9 (thỏa mãn)
Nếu 0 ≤ x < 4 thì (*) trở thành : x + 2 = 4 − x
⇔ x + x − 2 = 0 ⇔ x − 1 x + 2 = 0
Do x + 2 > 0 nên x = 1 ⇔ x = 1 (thỏa mãn)
Vậy có hai giá trị x=1 và x= 9 thỏa mãn yêu cầu bài toán
\(\left(-300\right):20+5.\left(3x-1\right)=25\)
\(\Leftrightarrow-15+5.\left(3x-1\right)=25\)
\(\Leftrightarrow5\left(3x-1\right)=40\)
\(\Leftrightarrow3x-1=8\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\)
Vậy x =3
(-15) + 5.[ 3x - 1] = 25
5.[ 3x - 1] = 25 - (-15)
5.[ 3x - 1] = 25 + 15
5.[ 3x - 1] = 40
3x - 1 = 40 : 5
3x - 1 = 8
3x = 8+ 1
3x = 9
x = 9:3
x =3
Vậy x = 3 HT nha