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Bài làm
x = 2+3+6+9+8+5+2+4+1+2123
x = ( 1+9 ) + ( 2+8 ) + ( 3+5+2 ) + ( 4+6 ) + 2123
x = 10 + 10 + 10 + 10 + 2123
x = 40 + 2123
x = 2163
Vậy x = 2163
# Chúc bạn học tốt #
=>3x(1-1/6)=3/4
=>3x=3/4:5/6=3/4*6/5=18/20=9/10
=>x=3/10
Đề là \(3x\left(\dfrac{1}{1}\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}\right)=\dfrac{3}{4}?\)
`3x (1/1 \times 1/2 + 1/2 \times 1/3 + 1/3 \times 1/4 + 1/4 \times 1/5 + 1/5 \times 1/6) = 3/4`
\(3x\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{3}{4}\)
\(3x\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{3}{4}\)
\(3x\left(1-\dfrac{1}{6}\right)=\dfrac{3}{4}\)
\(3x\times\dfrac{5}{6}=\dfrac{3}{4}\)
`3x=3/4 \div 5/6`
`3x = 9/10`
`x = 9/10 \div 3`
`x = 3/10.`
\(\dfrac{3}{5}\)x\(\dfrac{4}{7}:\dfrac{16}{21}\)
\(=\dfrac{12}{35}:\dfrac{16}{21}\)
\(=\dfrac{9}{20}\)
___________________________________
\(\dfrac{2}{3}-x=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}-\dfrac{2}{3}\)
\(x=0\)
Chúc bạn học tốt:>
\(a̸\)
\(\frac{3}{5}-\frac{1}{2}.x=\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{3}{5}-\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{7}{20}\)
\(\Rightarrow\frac{7}{10}\)
\(b̸\)
\(11,3+2\left[x-\frac{1}{3}\right]=\frac{25}{6}\)
\(2\left[x-\frac{1}{3}\right]=\frac{25}{6}-11,3\)
\(2\left[x-\frac{1}{3}\right]=\frac{-107}{15}\)
\(x-\frac{1}{3}=\frac{-107}{15}:2\)
\(x-\frac{1}{3}=\frac{-107}{30}\)
\(x=\frac{-107}{30}+\frac{1}{3}\)
\(x=\frac{-97}{30}\)
\(a)\frac{3}{5}-\frac{1}{2}x=\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{3}{5}-\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{7}{20}\)
\(\implies x=\frac{7}{20}:\frac{1}{2}\)
\(\implies x=\frac{7}{10}\)
Vậy...
\(b) 11,3+2(x-\frac{1}{3})=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-2.\frac{1}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-\frac{2}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x=\frac{25}{6}+\frac{2}{3}\)
\(\implies \frac{113}{10}+2x=\frac{29}{6}\)
\(\implies 2x=\frac{29}{6}-\frac{113}{10}\)
\(\implies 2x=\frac{-97}{15}\)
\(\implies x=\frac{-97}{30}\)
Vậy..
\(c)5x-435+2x+140+3x=565\)
\(\implies (5x+2x+3x)+(-435+140)=565\)
\(\implies 10x+(-295)=565\)
\(\implies 10x=565-(-295)\)
\(\implies 10x=860\)
\(\implies x=86\)
Vậy...
~ hok tốt a~
`@` `\text {Ans}`
`\downarrow`
`a)`
`5(x + 35) = 515`
`=> x + 35 = 515 \div 5`
`=> x+ 35 = 103`
`=> x = 103 - 35`
`=> x = 68`
Vậy, `x = 68`
`b)`
`12x - 33 = 3^2 * 3^3`
`=> 12x - 33 = 3^5`
`=> 12x = 3^5 + 33`
`=> 12x = 276`
`=> x = 276 \div 12`
`=> x = 23`
Vậy, `x = 23`
`c)`
`6x - 5 = 19`
`=> 6x = 19 + 5`
`=> 6x = 24`
`=> x = 24 \div 6`
`=> x = 4`
Vậy, `x = 4`
`d)`
`4(x - 12) + 9 = 17`
`=> 4(x - 12) = 17 - 9`
`=> 4(x - 12) = 8`
`=> x - 12 = 8 \div 4`
`=> x - 12 = 2`
`=> x = 14`
Vậy, `x = 14`
`e)`
123 - 5(x + 4) = 38`
`=> 5(x + 4)= 123 - 38`
`=> 5(x + 4) =85`
`=> x + 4 = 85 \div 5`
`=> x + 4 = 17`
`=> x = 13`
Vậy, `x = 13`
`f)`
`(3x - 2^4) * 7^3 = 2 * 7^4`
`=> 3x - 2^4 = 2 * 7^4 \div 7^3`
`=> 3x - 2^4 = 2* 7`
`=> 3x - 2^4 = 14`
`=> 3x = 14 + 2^4`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x = 10`
Vậy, `x = 10.`
a/+b/\(A\left(x\right)=2x^5+2-6x^2-3x^3+4x^5\)
\(=\left(2x^5+4x^5\right)-3x^3-6x^2+2\)
\(=6x^5-3x^3-6x^2+2\)
c/Bậc của \(A\left(x\right)\) là 5
d/\(A\left(1\right)=6\cdot1^5-3\cdot1^3-6\cdot1^2+2\)
\(=6-3-6+2\)
\(=-1\)
\(A\left(-2\right)=6\cdot\left(-2\right)^5-3\cdot\left(-2\right)^3-6\cdot\left(-2\right)^2+2\)
\(=6\cdot\left(-32\right)-3\cdot\left(-8\right)-6\cdot4+2\)
\(=-192-\left(-24\right)-24+2\)
\(=-190\)
a) và b)
A(x) = 2x⁵ + 2 - 6x² - 3x³ + 4x⁵
= (2x⁵ + 4x⁵) - 3x³ - 6x² + 2
= 6x⁵ - 3x³ - 6x² + 2
c) Bậc của A(x) là 5
d) A(1) = 6.1⁵ - 3.1³ - 6.1² + 2
= 6.1 - 3.1 - 6.1 + 2
= 6 - 3 - 6 + 2
= -1
A(2) = 6.2⁵ - 3.2³ - 6.2² + 2
= 6.32 - 3.8 - 6.4 + 2
= 192 - 24 - 24 + 2
= 146
a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)
\(\Leftrightarrow3x+2x-10=6-5x+1\)
\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm
b, \(x^3-3x^2-x+3=0\)
\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)
Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }
c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3+x^2-3x-2=0\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn
Vậy ...
Không cho đề bài thì biết làm kiểu gì
Em thêm đề bài nha