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Tranthi Tiu
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Nguyễn Lê Phước Thịnh
1 tháng 12 2022 lúc 0:07

a: \(=12x^2-9x-12x^2-10x+6x+5=-13x+5\)

b: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x\)

\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)

\(=x^3-2x^2+3x\)

c: \(=x^3-3x^2+3x-1+x^3+8+3\left(x^2-16\right)\)

\(=2x^3-3x^2+3x+7+3x^2-48=2x^3+3x-41\)

d: \(=\left(x^3+1\right)\left(x^3-1\right)=x^6-1\)

hoang thi an
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Phạm Ngân Hà
1 tháng 8 2018 lúc 20:37

a) \(-2\left(x-\dfrac{1}{3}\right)-5\left(x+\dfrac{1}{3}\right)=\dfrac{1}{2}x\)

\(\Leftrightarrow-2x+\dfrac{2}{3}-5x-\dfrac{5}{3}=\dfrac{1}{2}x\)

\(\Leftrightarrow-7x-1=\dfrac{1}{2}x\)

\(\Leftrightarrow-7x-\dfrac{1}{2}x=1\)

\(\Leftrightarrow-\dfrac{15}{2}x=1\)

\(\Leftrightarrow x=1:\left(-\dfrac{15}{2}\right)=-\dfrac{2}{15}\)

b) \(-\left(\dfrac{1}{2}x-\dfrac{3}{4}\right)-\left(-x+1\right)=\dfrac{3}{2}\)

\(\Leftrightarrow-\dfrac{1}{2}x+\dfrac{3}{4}+x-1=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{2}+\dfrac{1}{4}=\dfrac{7}{4}\)

\(\Leftrightarrow x=\dfrac{7}{4}:\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

Linh Luna
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Nguyễn Lê Phước Thịnh
28 tháng 5 2022 lúc 22:27

a: \(\left|\dfrac{1}{2}x-3\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=\dfrac{1}{2}\\\dfrac{1}{2}x-3=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{7}{2}\\\dfrac{1}{2}x=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)

b: \(\left|3x+\dfrac{1}{5}\right|-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow\left|3x+\dfrac{1}{5}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{5}=1\\3x+\dfrac{1}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{4}{5}\\3x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{15}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: \(\left|\dfrac{1}{2}x-3\right|< \dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3>-\dfrac{1}{2}\\\dfrac{1}{2}x-3< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x>\dfrac{5}{2}\\\dfrac{1}{2}x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow5< x< 7\)

e: \(\left|3x-\dfrac{4}{5}\right|>\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}>\dfrac{1}{3}\\3x-\dfrac{4}{5}< -\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>\dfrac{17}{15}\\3x< \dfrac{7}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{17}{45}\\x< \dfrac{7}{45}\end{matrix}\right.\)

Thảo Anh Trần Ngọc
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Thảo Anh Trần Ngọc
7 tháng 3 2020 lúc 22:05

MỌI NGƯỜI GIÚP MÌNH VỚI Ạ. AI NHANH MÌNH TICK NHA

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Mai Tran
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NT Mai Hương
21 tháng 8 2017 lúc 9:38

a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0

Khánh Linh
22 tháng 8 2017 lúc 21:19

a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran

Khánh Linh
22 tháng 8 2017 lúc 21:25

c, Do (x - \(\dfrac{1}{2}\))(x + \(\dfrac{3}{4}\)) < 0
=> x - \(\dfrac{1}{2};x+\dfrac{3}{4}\) khác dấu
Có x - \(\dfrac{1}{2}< x+\dfrac{3}{4}\Rightarrow x-\dfrac{1}{2}< 0;x+\dfrac{3}{4}>0\)
=> \(\left[{}\begin{matrix}x-\dfrac{1}{2}< 0\\x+\dfrac{3}{4}>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>-\dfrac{3}{4}\end{matrix}\right.\)<=> -\(\dfrac{3}{4}< x< \dfrac{1}{2}\)
Do x \(\in Z\Rightarrow x=0\)
@Mai Tran

French Fries Mlem Mlem
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Lê Trang
2 tháng 9 2020 lúc 21:34

Bài 3: Tìm x, biết:

a) \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow4x=40\)

\(\Leftrightarrow x=10\)

Vậy x = 10

b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow\left(2x+3\right)^2-4\left(x^2-1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)

\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18\)

\(\Leftrightarrow2\left(1-2x\right)=18\)

\(\Leftrightarrow2-4x=18\)

\(\Leftrightarrow4x=-16\)

\(\Leftrightarrow x=-4\)

Vậy x =-4

d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

\(\Leftrightarrow x=\frac{5}{12}\)

Vậy \(x=\frac{5}{12}\)

e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow25-6x=9\)

\(\Leftrightarrow6x=16\)

\(\Leftrightarrow x=\frac{8}{3}\)

Vậy \(x=\frac{8}{3}\)

f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow21-5x=0\)

\(\Leftrightarrow5x=21\)

\(\Leftrightarrow x=\frac{21}{5}\)

Vậy \(x=\frac{21}{5}\)

Lê Trang
2 tháng 9 2020 lúc 21:47

bài của bạn làm sai rồi :)

Lê Đức Minh Nhật
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Tùng Hà
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💋Amanda💋
25 tháng 3 2020 lúc 20:08
https://i.imgur.com/9YAi91S.jpg
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nguyễn thi trà giang
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qwerty
8 tháng 8 2017 lúc 7:47

đề "mờ ám" quá