Tìm \(x\in Z\) sao cho:
\(\left(17x-7\right)+\left(17x-6\right)+\left(17x-5\right)+...+\left(7x-19\right)+\left(7x-18\right)+\left(7x-17\right)=-14544\)
Những câu hỏi liên quan
LÀM TÍNH CHIA:
a) \(\left(6x^6+2x^5-2x^4-15x^3+x^2+7x+2\right):\left(3x^2+x-1\right)\)
b) \(\left(-6x^4+5x^3+17x^2-23x+7\right):\left(-3x^2-2x+7\right)\)
Mọi người giải giúp mình bài này với:B2: Giải các PT sau:l) left(x-1right)left(5x+3right)left(3x-8right)left(x-1right)m) 2xleft(x-1right)x^2-1n) left(2-3xright)left(x+11right)left(3x-2right)left(2-5xright)o) frac{3}{7}x-1frac{1}{7}xleft(3x-7right)p) left(x-frac{3}{4}right)^2+left(x-frac{3}{4}right)left(x-frac{1}{2}right)0q) frac{1}{x}x+2left(frac{1}{x}+2right)left(x^2+1right)r) left(2x+3right)left(frac{3x+8}{2-7x}+1right)left(x-5right)left(frac{3x+8}{2-7x}+1right)s) left(x+2right)left(x-3right)l...
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Mọi người giải giúp mình bài này với:
B2: Giải các PT sau:
l) \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
m) \(2x\left(x-1\right)=x^2-1\)
n) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
o) \(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)
p) \(\left(x-\frac{3}{4}\right)^2+\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}\right)=0\)
q) \(\frac{1}{x}x+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
r) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
s) \(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
Các Pro giúp mình với !
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
Xem thêm câu trả lời
\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
\(\left(x+2\right)\left(x-3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x-3\right)\left(x^2-17x+33\right)\)
=>\(17x^2-17x+8=x^2-17x+33\)
<=> \(16x^2-25=0\)
<=>\(\left(4x-5\right)\left(4x+5\right)=0\)
=> \(4x-5=0=>x=\dfrac{5}{4}\)
hoặc \(4x+5=0=>x=\dfrac{-5}{4}\)
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(x+2)(x−3)(17x2−17x+8)=(x+2)(x−3)(x2−17x+33)
\(\Leftrightarrow\)(x+2)(x−3)(17x2−17x+8) - (x+2)(x−3)(x2−17x+33) = 0
\(\Leftrightarrow\)(x+2)(x−3).[(17x2−17x+8)-(x2−17x+33)] = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\text{x+2 = 0}\\\text{x−3 = 0}\\\text{(17x^2−17x+8)-(x^2−17x+33) = 0}\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=3\\17x^2-17x+8-x^2+17x-33=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\16x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\\left(4x-5\right)\left(4x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x-5=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\4x=5\\4x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{5}{4}\\x=\dfrac{-5}{4}\end{matrix}\right.\)
Vậy S = \(\left\{-2;\dfrac{-5}{4};\dfrac{5}{4};3\right\}\)
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\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}=3\sqrt{3}\left(x+2\right)\)
Ta có:
\(\sqrt{x^2-x+19}+\sqrt{7x^2+8x+13}+\sqrt{13x^2+17x+7}\)
\(=\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{75}{4}}+\sqrt{\left(2x-1\right)^2+3\left(x+2\right)^2}+\sqrt{\frac{1}{4}\left(2x-1\right)^2+\frac{3}{4}\left(4x+3\right)^2}\)
\(\ge\sqrt{\frac{75}{4}}+\sqrt{3\left(x+2\right)^2}+\sqrt{\frac{3}{4}\left(4x+3\right)^2}\)
\(=\frac{5\sqrt{3}}{2}+\sqrt{3}\left(x+2\right)+\frac{\sqrt{3}\left(4x+3\right)}{2}=3\sqrt{3}\left(x+2\right)\)
Dấu = xảy ra khi ....
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a)\(\left|5x+4\right|+7=26\)
b)\(3\left|9-2x\right|-17=16\)
c)\(8x-\left|4x+1\right|=x+2\)
d)\(\left|17x-5\right|-\left|17x+5\right|=0\)
e)\(\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|=2\)
Tìm X
a: =>|5x+4|=19
=>5x+4=19 hoặc 5x+4=-19
=>5x=15 hoặc 5x=-23
=>x=3 hoặc x=-23/5
b: =>3|2x-9|=33
=>|2x-9|=11
=>2x-9=11 hoặc 2x-9=-11
=>2x=20 hoặc 2x=-2
=>x=10 hoặc x=-1
d: =>|17x-5|=|17x+5|
=>17x-5=17x+5 hoặc 17x-5=-17x-5
=>34x=0
hay x=0
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218. Tìm x biết:
a) \(\left|17x-5\right|-\left|17x+5\right|=0\)
b) \(\left|3x+4\right|=2.\left|2x-9\right|\)
Nhầm sorry mk tưởng cộng sory bạn nha Thái Viết Nam
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Ta có : |17x - 5| - |17x + 5| = 0
Mà |17x - 5| \(\ge\)0 ; |17x + 5| \(\ge\) 0
Nên \(\hept{\begin{cases}\left|17x-5\right|=0\\\left|17x+5\right|=0\end{cases}}\)
<=>\(\hept{\begin{cases}17x-5=0\\17x+5=0\end{cases}}\)
<=> \(\hept{\begin{cases}17x=5\\17x=-5\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{5}{17}\\x=-\frac{5}{17}\end{cases}}\)
Mà x ko thể đồng thời bằng 2 giá trị
Nên x thuộc rỗng
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Tìm x biết1) left(2x+3right)left(x-4right)+left(x-5right)left(x-2right)left(3x-5right)left(x-4right)2)left(8x-3right)left(3x+2right)-left(4x+7right)left(x+4right)left(2x+1right)left(5x+1right)-333)6xleft(3x+5right)-2xleft(9x-2right)+left(17-xright)left(x-1right)+xleft(x-18right)-17x^204)left(x-1right)left(x+2right)-left(x-3right)+5x-70Giúp mình nha. Camon nhiều
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Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
tính x biết:\(\left|17x-5\right|-\left|17x+5\right|=0\)
<=> |17x - 5| = |17x + 5|
=> 17x - 5 = 17x + 5 hoặc 17x - 5 = -17x - 5
=> 0x = 10(loại) hoặc 34x = 0
<=> x = 0.
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pt <=> | 17x - 5 | = | 17x + 5 |
\(\Leftrightarrow\orbr{\begin{cases}17x-5=17x+5\left(loai\right)\\17x-5=-17x-5\end{cases}}\)
\(\Leftrightarrow17x=-17x\)
\(\Leftrightarrow x=-x\)
\(\Leftrightarrow x=0\)
tìm x
a) \(2\left|x-6\right|+7x-2=\left|x-6\right|+7x\)
b)\(4\left(x-5\right)-7\left(5-x\right)+10\left(5-x\right)=-3\)
\(a.\Leftrightarrow2|x-6|-|x-6|-2=0\)
\(\Leftrightarrow|x-6|-2=0\)
\(\Leftrightarrow|x-6|=2\)
\(+x-6=2\)
\(\Leftrightarrow x=8\)
\(+x-6=-2\)
\(\Leftrightarrow x=4\)
v...
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\(b.\Leftrightarrow-4\left(5-x\right)-7\left(5-x\right)+10\left(5-x\right)=-3\)
\(\Leftrightarrow\left(5-x\right)\left(10-4-7\right)=-3\)
\(\Leftrightarrow-1.\left(5-x\right)=-3\)
\(\Leftrightarrow5-x=3\)
\(\Leftrightarrow x=2\)
v...
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