\(20\%\times x+x=42\)
k mk nha
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Ta có: \(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}=42^5:\left(2^3\cdot21^6\right)\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5\cdot21^5}{2^3\cdot21^5\cdot21}\)
\(\Leftrightarrow\left(x-2\right)\cdot\frac{4}{21}=\frac{4}{21}\)
\(\Leftrightarrow x-2=1\)
hay x=3
Vậy: x=3
a, x ( x - 2 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b, x : 15 + 42 = 15 + 25 . 8
x : 15 + 42 = 15 + 200
x : 15 + 42 = 215
x : 15 = 215 - 42
x : 15 = 173
x = 173 . 15
x 2595
x.(x-2)=0
=> x=0 hoặc x-2=0
x:15+42=15+25.8
x:15+42=215
x:15=215-42
x:15=173
x=173.15
x=2595
a. x + 2,5 - 1,5 = 20 x+2,5 = 20 + 1,5 x+ 2,5=21,5 x= 21,5 - 2,5 x= 19 b. x - 10 + 5 = 10 x - 10 = 10 - 5 x - 10 = 5 x = 5 + 10 x = 15
a) x+2,5-1,5=10x2
x+(2,5-1,5)=20
x+1=20
x = 20-1
x = 19
b) x-(10+5)=(20-10)
x -15=10
x = 15+10
x = 25
nhớ k minh minh k lại
2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 42 + 27 )
= 24 x 100
= 2400
a) A = a + (42-70+18) - (42+18+a)
= a + 42 - 70 + 18 - 42 - 18 - a
= (a-a) + (42-42) + (18-18) - 70
= 0 + 0 + 0 - 70 = -70.
Vậy A = -70.
b) B = a + 30 + 12 - (-20) + (-12) - (2+a)
= a + 30 + 12 + 20 - 12 - 2 - a
= (a-a) + (12-12) + (30+20-2)
= 0 + 0 + (50-2)
= 50 - 2 = 48.
Vậy B = 48.
c) C = (x-y+z) - (x-y-z) - (2x+y)
= x - y + z - x + y + z - 2x - y
= (x-x-2x) + (-y+y-y) + (z+z)
= -2x + (-y) + 2z
= -2x - y + 2z.
Vậy C = -2x - y + 2z.
a)
\(x^2-5x+4x-20=0.\)
\(x^2-x-20=0\)
\(\left(x^2-x+\frac{1}{4}\right)-20-\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2-\left(\frac{20.4+1}{4}\right)=0\)
\(\hept{\begin{cases}x-\frac{1}{2}-\left(\frac{20.4+1}{4}\right)=0\\x-\frac{1}{2}+\left(\frac{20.4+1}{4}\right)=0\end{cases}}\)
b) \(x^2+6x-7x-42=0\)
\(x^2-x-42=0\)
\(x^2-x+\frac{1}{4}-42-\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2-\left(\frac{42.4+1}{4}\right)=0\) " tương tự con A
\(x^3-16x=0\)
\(x\left(x^2-16\right)=0\)
\(x=0,+4,-4\)
\(x^3-16x=0\)
\(x.\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=16\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)
Vậy \(x=0\)hoặc \(x=\pm4\)
Tham khảo nhé~
Ta có: A=(1-1/2)...........................
Mà các tử có hiệu bằng 0
suy ra: Phân số có tử bằng 0
suy ra: A=0
Vậy A=0
\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}=1\)
\(x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=1\)
\(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)
\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=1\)
\(x-\left(1-\frac{1}{7}\right)=1\)
\(x-\frac{6}{7}=1\)
\(x=1+\frac{6}{7}\)
\(x=\frac{7}{7}+\frac{6}{7}=\frac{13}{7}\)
Vậy \(x=\frac{13}{7}\)
20%*x+x=42
x(20%+1)=42
x(1/5+1)=42
x*1,2=42
x=42/1,2=35
20%.x+x=42
x(20%+1)=42
x.1,2=42
x=42/1,2
x=35