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\(a)24\times(x-16)=12^2\\\Rightarrow 24\times(x-16)=144\\\Rightarrow x-16=144:24\\\Rightarrow x-16=6\\\Rightarrow x=6+16\\\Rightarrow x=22\\---\\b)(x^2-10):5=5\\\Rightarrow x^2-10=5\times5\\\Rightarrow x^2-10=25\\\Rightarrow x^2=25+10\\\Rightarrow x^2=35\\\Rightarrow x=\pm\sqrt{35}\\---\)
\(c)(5x+335):2=400\\\Rightarrow 5x+335=400\times2\\\Rightarrow 5x+335=800\\\Rightarrow 5x=800-335\\\Rightarrow 5x=465\\\Rightarrow x=465:5\\\Rightarrow x=93\\---\\d)63:(5x+4)=2^3-1\\\Rightarrow 63:(5x+4)=8-1\\\Rightarrow 63:(5x+4)=7\\\Rightarrow 5x+4=63:7\\\Rightarrow 5x+4=9\\\Rightarrow 5x=9-4\\\Rightarrow 5x=5\\\Rightarrow x=5:5\)
\(\Rightarrow x=1\)
\(Toru\)
Bài 1:
a) Ta có: \( \left|x+2\right|=\left|3-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=3-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=-3-2\\x+2x=3-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-5\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left|2x-4\right|=\left|3-x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=3-x\\2x-4=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=3+4\\2x-x=-3+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=7\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=1\end{matrix}\right.\)
c) C=(151515/161616 + 17^9/17^10)-(1500/1600 - 1616/1717)
=(15/16 + 1/17)-(15/16 - 16/17)
= 15/16 ( 1/17 + 16/17)
=15/16 . 1 = 15/16
Câu 13:
a) 50%x=-3/4
x=-3/4:50%
x=-3/2
(1,4x-16):2/3=-45
1,4x-16 =-45.2/3
1,4x-16 =-30
1,4x =-30+16
1,4x =-14
x =-14:1,4
x =-10
b)A=[1 13/15.(-13/97-53/-79)-13/15:-79/53].(1/3-0,25-3/4.1/9)/ 2/5+5+1,75
A=[1 13/15.(-13/97-53/-79)-13/15:-79/53].(1/3-0,25-1/12)/ 2/5+5+1,75
A=[1 13/15.(-13/97-53/-79)-13/15:-79/53].0/ 2/5+5+1,75
A=0/ 2/5+5+1,75
A=0
Câu 14:
Giải
a) Số h/s của lớp 6A là:
16:2/5=40 (h/s)
b) Số h/s giỏi của lớp 6A là:
40.3/20=6 (h/s)
Số h/s khá của lớp 6A là:
40-(16+6)=18 (h/s)
câu a;
6.(\(x+11\)) - 7.(2 - \(x\)) = 26
6\(x\) + 66 - 14 + 7\(x\) = 26
(6\(x\) + 7\(x\)) + (66 - 14) = 26
13\(x\) + 52 = 26
13\(x\) = 26 - 52
13\(x\) = - 26
\(x\) = - 26 : 13
\(x\) = - 2
a: \(6\left(x+11\right)-7\left(2-x\right)=26\)
=>6x+66-14+7x=26
=>13x+52=26
=>13x=-26
=>x=-26:13=-2
b: \(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}-\dfrac{x+21}{2023}-\dfrac{x+20}{2024}=0\)
=>\(\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)-\left(\dfrac{x+21}{2023}+1\right)-\left(\dfrac{x+20}{2024}+1\right)=0\)
=>\(\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}-\dfrac{x+2024}{2023}-\dfrac{x+2024}{2024}=0\)
=>\(\left(x+2044\right)\left(\dfrac{1}{2021}+\dfrac{1}{2022}-\dfrac{1}{2023}-\dfrac{1}{2024}\right)=0\)
=>x+2044=0
=>x=-2044
c: \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=120x\)
mà \(VT>=0\forall x\)
nên 120x>=0
=>x>=0
=>\(\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2};\left|x+\dfrac{1}{2\cdot3}\right|=x+\dfrac{1}{6};...;\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{99\cdot100}\)
Phương trình sẽ tương đương với:
\(x+\dfrac{1}{2}+x+\dfrac{1}{2\cdot3}+...+x+\dfrac{1}{99\cdot100}=120x\)
=>\(100x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=120x\)
=>20x=1-1/100=99/100
=>\(x=\dfrac{99}{2000}\)