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\(\left(5x-29\right)-\left(2x-29\right)=-21\)
\(5x-29-2x+29=-21\)
\(5x-2x=-21+29-29\)
\(3x=-21\)
\(x=-7\)
a)15/16:3/8x3/4
=15/16:6/16x12/16
=15/16x16/6x12/16
=15/6x12/16
=30/16=15/8
b)5/11x18/29-5/11x8/29+5/11x19/29
=5/11x(18/29-8/29+19/29)
=5/11x1=5/11
a) x+(-45)=(-62)+17
=>x-45=-45
=>x=-45+45
=>x=0
b) x+29=/-43/+(-43)
=>x+29=43-43
=>x+29=0
=>x=-29
c) 5+/x+1/=29
=>/x+1/=29-5=24
\(\orbr{\begin{cases}x+1=24\\x+1=-24\end{cases}}\Rightarrow\orbr{\begin{cases}x=23\\x=-25\end{cases}}\)
P/s tham khảo bài mình nha
a ) x + (-45) =-45
= > x = 0
b ) x + 29 = 0
=> x = -29
c ) => /x+1/ = 29-5
/x+1/ = 24
=> x+1 = + - 24
TH1 : x+1=24
=> x =23
th2 : x+1 = -24
=> x =-25
vậy x thuộc { 23 ; -25}
a,( 3/29 - 1/5 ) x 29/3
=(15/145-29/145)x29/3
=(-14/145)x29/3
=-406/435
=-14/15
b,4 x 11 x 3/4 x 9/121
=4 x 11 x 3x9/1x1x4x121
=1188/484
=59/242
k mk nha
A / ( 3/29 - 1/5 ) x 29/3 = -14/145 x 29/3 = -14/15 B/ 4 x 11 x 3/4 x 9/121 = (4 x 11) x ( 3/4 x 9/121 ) = 44 x 27/484 =27/11 xong rồi đó
dau . la nhan nha em
a) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{29}{3}.\frac{3}{29}-\frac{1}{5}.\frac{29}{3}=1-\frac{29}{15}=\frac{-14}{15}\)
b)\(4.11.\frac{3}{4}.\frac{9}{121}\)
\(=\left(4.\frac{3}{4}\right).\left(11.\frac{9}{121}\right)\)
\(=3.\frac{9}{12}\)
\(=\frac{9}{4}=2\frac{1}{4}\)
A=((2-2/19+2/23-1/1010)/(3-3/19+3/23-3/2020))x((4-4/29+4/41-1/505)/(5-5/29+5/41-1/404))
giúp mình với
\(1,-\frac{3}{29}+\frac{-7}{29}\le\frac{x}{29}\le-\frac{3}{29}-\frac{5}{29}\)
\(\Rightarrow-\frac{10}{29}\le\frac{x}{29}\le-\frac{8}{29}\Rightarrow-10\le x\le-8\)
\(\Rightarrow x=\left\{-8;-9;-10\right\}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S-S=S=1-\frac{1}{2^{100}}\)
a) \(5-\left[x+1\right]=29\)
\(x+1=5-29\)
\(x+1=-24\)
\(x=-24-1\)
\(x=-25\)
Vậy \(x=-25\) là giá trị cần tìm
b) \(\left(-1\right)+3+\left(-5\right)+7+\left(-9\right)+........+x=600\)
\(\Rightarrow\left[\left(-1\right)+3\right]+\left[\left(-5\right)+7\right]+............+\left[-\left(x-2\right)+x\right]=600\)
\(\Rightarrow2+2+2+.......+2=600\)
\(\Rightarrow2\left(1+1+......+1\right)=600\)
\(\Rightarrow1+1+........+1=300\)
Số dấu ngoặc [] là :
\(\dfrac{x-3}{4+1}\)
\(\Rightarrow\dfrac{x-3}{4+1}=300\)
\(\Rightarrow\dfrac{x-3}{4}=299\)
\(\Rightarrow x-3=299.4\)
\(\Rightarrow x=299.4+3\)
\(\Rightarrow x=1199\)
\(5-\left|x+1\right|=29\)
\(\left|x+1\right|=5-29\)
\(\left|x+1\right|=-24\Leftrightarrow x\in\left\{\varnothing\right\}\)
=>/x+1/=29-5
=> / x+1 /= 24
=>hoặc x+1 = 24
hoặc x+1= -24
=> hoặc x=24-1=23
hoặc x= -24 - 1 = -25
VẬY x thuộc{23; -25}
tk mình đi