5x+2-5x=105 ;204-4(x+1)=43
Hãy tìm x
105 + | 3x - 7 | + | 5x - 2 | = 405
a) 16+|2x+5|-|x-1|=5x+9
b) 105+|3x-7|+|5x-2=405
c) 302-|x+4|-|3x-7|=9
d) 415+|4-x|+|8x-2|=515
Tìm giá trị của đa thức sau:
\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)biết \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)
\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)
\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)
\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)
Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:
\(E=5x.0+105=105\)
Tìm giá trị của đa thức sau:
\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\) biết \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)
5x-3-20=105
ai bt làm bài tìm x này giúp mik với
trả lời
5x-3 -20 = 105
5x-3 = 105 + 20
5x-3 = 125
5x-3 = 53
x-3 = 3
x =3+3
x = 6
Vậy x = 6
\(\frac{5x}{2}=\frac{6y}{5}=\frac{7z}{3},y-x-z=\frac{1}{105}\)
Ta có: \(\frac{5x}{2}=\frac{6y}{5}=\frac{7z}{3}\) => \(\frac{x}{\frac{2}{5}}=\frac{y}{\frac{5}{6}}=\frac{z}{\frac{3}{7}}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{\frac{2}{5}}=\frac{y}{\frac{5}{6}}=\frac{z}{\frac{3}{7}}=\frac{y-x-z}{\frac{5}{6}-\frac{2}{5}-\frac{3}{7}}=\frac{\frac{1}{105}}{\frac{1}{210}}=2\)
=> \(\hept{\begin{cases}\frac{x}{\frac{2}{5}}=2\\\frac{y}{\frac{5}{6}}=2\\\frac{z}{\frac{3}{7}}=2\end{cases}}\) => \(\hept{\begin{cases}x=\frac{4}{5}\\y=\frac{5}{3}\\z=\frac{6}{7}\end{cases}}\)
Tìm số tự nhiên x biết:
a) 71 . 2 - 6 ( 2 x + 5 ) = 10 5 : 10 3
b) ( 5 x + 3 4 ) . 6 8 = 6 9 . 3 4
tìm các cặp số tự nhiên x, y biết:
xy+5x+5y=92xy+5x-2y=105xy-3x+2y-11=0xy-2x+y-4=0xy-x+2y-2=42xy+2x+3y=12nhanh minh tick
1. xy + 5x + 5y = 92
=> (xy + 5x) + (5y + 25) = 92 + 25
=> x(y + 5) + 5(y + 5) = 117
=> (x + 5)(y + 5) = 117
=> x + 5 \(\in\)Ư(117) = {-1;1;-3;3;-9;9;-13;13;-39;39;-117;117}
Mà x >= 0 => x + 5 >= 5
=> x + 5 \(\in\){9;13;39;117}
Ta có bảng sau:
x + 5 | 9 | 13 | 39 | 117 |
x | 4 | 8 | 34 | 112 |
y + 5 | 13 | 9 | 3 | 1 |
y | 8 | 4 | -2 (loại) | -4 (loại) |
Vậy; (x;y) \(\in\){(4;8);(8;4)}
các câu còn lại tương tự như bài mình vừa làm
1) 2(3x+5)-6=9
2) 5x+3(4+2x)=25
3) 3(4x+1)+2(x-1)=105
4) 30-[2(x-3)-2]=14
1) 2(3x+5)-6=9
2(3x+5) =9+6
2(3x+5) =15
3x+5 = 15:2
3x+5 = \(\frac{15}{2}\)
3x = \(\frac{15}{2}-5\)
3x =\(\frac{5}{2}\)
x = \(\frac{5}{2}:3\)
x = \(\frac{5}{6}\)
2) 5x+3(4+2x)=25
5x+12+6x=25
5x+6x =-12+25
11x =13
x =13:11
x =\(\frac{13}{11}\)
3) 3(4x+1)+2(x-1)=105
12x+3+2x-2=105
12x+2x = -3+2+105
14x =104
x = 104:14
x = \(\frac{52}{7}\)
4) \(30-[2\left(x-3\right)-2]=14\)
\(2\left(x-3\right)-2\)=\(30-14\)
\(2\left(x-3\right)-2=16\)
\(2\left(x-3\right)=16+2\)
\(2\left(x-3\right)=18\)
\(x-3=18:2\)
\(x-3=9\)
\(x=9+3\)
\(x=12\)
k mình nha !!!!!!
(2x-6)(x2+2)=(2x-6)(8x-10)
(5x-1)2=(3x+5)2
109-x/91 +107-x/93 +105-x/95+103-x/97=-4
\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)
\(\Rightarrow x\in\left\{3;6;2\right\}\)
\(\left(5x-1\right)^2=\left(3x+5\right)^2\)
\(\Leftrightarrow\left(5x-1\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(5x-1-3x-5\right)\left(5x-1+3x+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-4\)
\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\Leftrightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow200-x=0\)
\(\Leftrightarrow x=200\)
Vậy....