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Bài 1:
a) 2/3x + 1/2 = 1/10
2/3x = -2/5
x= -3/5
b) 2/3x + 1/5 = 7/10
2/3x= 1/2
x= 3/4
c) \(\left(3\frac{4}{5}-2x\right)\cdot1\frac{1}{3}=5\frac{5}{7}\)
\(\left(\frac{19}{5}-2x\right)\cdot\frac{4}{3}=\frac{40}{7}\)
\(\frac{19}{5}-2x=\frac{30}{7}\)
\(2x=-\frac{17}{35}\)
\(x=-\frac{17}{70}\)
d) x/7 = 6/-21
=> -21x = 42
x= -2
Bài 2:
a) [3/8 +(-1/4) + 5/2] : 2/3
= (3/8 - 2/8 + 20/8) :2/3
= 21/8 : 2/3
= 63/16
b) -5/7 * 2/11 + (-5/7) *9
= -10/77 - 45/7
= 485/77
c) 0,25 : (10,3 - 9,8) -3/4
= 0,25 : 0,5 - 0,75
= 0,5 - 0,75
= -0,25
d) \(\frac{3}{4}+\frac{1}{5}:\left(-\frac{7}{10}\right)\)
= 3/4 - 2/7
= 13/28
Cậu vui lòng kiểm tra lại đề trước khi xem phần giải ở dưới mỗi câu nhé. Tại tớ sợ tớ chép sai đề.
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
Bài 1 :Bỏ dấu ngoặc
2007-(7-3+4)
= 2007 -7+3-4
= 1999
6+[(-5) + 4 - 1 ]
= 6-5+4-1
=4
5-[(-6+8-2]
= 5+6-8+2
=5
-10+(7-3+1)
= -10 +7-3+1
= -5
Bài 3 Tìm x
\(\dfrac{1}{3} = \dfrac{x}{6}\)
\(<=> x= \dfrac{1.6}{3}\)
\(<=> x=2\)
\(a,\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+\frac{17}{-35}\)
=\(-\frac{5}{13}+\left(\frac{3}{17}+\frac{14}{17}\right)+\left(\frac{-18}{35}+\frac{-17}{35}\right)\)
= \(-\frac{5}{13}+1+\left(-1\right)\)
=\(-\frac{5}{13}\)
\(b,\frac{-3}{8}.\frac{1}{6}+\frac{3}{-8}.\frac{5}{6}+\frac{-10}{6}\)
=\(\frac{-3}{8}.\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{-10}{6}\)
=\(\frac{-3}{8}.1+\frac{-10}{6}\)
=\(-\frac{49}{24}\)
\(c,\frac{-4}{11}.\frac{5}{15}.\frac{11}{-4}\)
=\(\left(\frac{-4}{11}.\frac{11}{-4}\right).\frac{1}{3}\)
=\(1.\frac{1}{3}=\frac{1}{3}\)
\(d,\frac{13}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}.\frac{1}{2}\)
=\(\frac{13}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}\)
=\(\frac{13}{8}+\frac{1}{2}+\frac{-1}{8}\)
=\(\left(\frac{13}{8}+\frac{-1}{8}\right)+\frac{1}{2}\)
=\(\frac{3}{2}+\frac{1}{2}=2\)
\(e,\frac{-1}{2^2}-\left(-2\right)^2-5\)
=\(\frac{-1}{4}-4-5\)
=\(-\frac{37}{4}\)
\(f,\frac{121}{3}-\frac{5}{7}:\left(24-\frac{23}{57}\right)\)
=\(\frac{121}{3}-\frac{5}{7}:\frac{1345}{57}\)
=\(\frac{121}{3}-\frac{57}{1883}\)
\(\approx40,4\)
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
Nên đợi ai đó giải hết 2 3 bài xong rồi mới đăng tiếp những bài còn lại, chứ dài vậy giải hơi nản =)))
Bài 1:
1, \(13\frac{2}{5}-\left(\frac{18}{32}-2\frac{6}{10}\right)\)
\(=\frac{67}{5}-\left(\frac{9}{16}-\frac{13}{5}\right)\)(Chuyển hỗn số thành p/số và rút gọn hai số trong ngoặc luôn)
\(=\frac{67}{5}-\left(\frac{-163}{80}\right)\)
\(=\frac{246}{16}\)
2, \(22.4\frac{5}{7}-\left(8.91+1,09\right)\)(Phần 2 viết vầy có đúng không vậy ? Nếu sai thì kêu chị sửa nhé)
\(=22.\frac{33}{7}-10\)
\(=\frac{726}{7}-10\)
\(=\frac{656}{7}\)
3, Chỗ ''3 phần 10 phần 2'' là sao :v ?
4, \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
\(=\frac{37}{7}.\frac{8}{11}+\frac{37}{7}.\frac{5}{11}-\frac{37}{7}.\frac{2}{11}\)(Chuyển hỗn số thành p/số)
\(=\frac{37}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)(Dùng tính chất phân phối)
\(=\frac{37}{7}.\frac{11}{11}\)
\(=\frac{37}{7}.1=\frac{37}{7}\)