/x+1/+/x+2/+/x+3/+/x+4/=5x-1
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a, Có : |2x-1| + |2x-5| = |2x-1| + |5-2x| >= |2x-1+5-2x| = 4
Dấu "=" xảy ra <=> (2x-1).(5-2x) = 0 <=> 1/2 < = x < = 5/2
Vậy 1/2 < = x < = 5/2
b, +, Nếu x < -3/4 => 3-x-3x-4 = -2x-1
=> ko tồn tại x
+, Nếu -3/4 < = x < = -1/2 => 3-x+3x+4 = -2x-1
=> x=-2 ( ko t/m )
+, Nếu -1/2 < x < = 3 => 3-x+3x+4=2x+1
=>ko tồn tại x
+, Nếu x > 3 => x-3+3x+4=2x+1
=> x=0 ( ko t/m )
Vậy ko tồn tại x t/m bài toán
Tham khảo xem có đúng ko nha !
Tính phải k nhỉ?
`1)`
`2x + 3x + 5x`
`= (2 + 3 + 5)x`
`= 10x`
`2)`
`2.x - x + 3.x`
`= (2 - 1 + 3)x`
`= 4x`
`3)`
`9.x - 3 - 3.x`
`= (9 - 3)x - 3`
`= 6x - 3`
`4)`
Thiếu dấu, bạn bổ sung thêm
`5)`
`x - 0,2x - 0,1x`
`= (1 - 0,2 - 0,1)x`
`=0,7x`
`6)`
\(\dfrac{7}{2}x-\dfrac{1}{2}x=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)x=3x\)
`@` `\text {Ans}`
`\downarrow`
`1)`
\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(2x=\dfrac{7}{6}\)
`\Rightarrow`\(x=\dfrac{7}{6}\div2\)
`\Rightarrow`\(x=\dfrac{7}{12}\)
Vậy, `x = 7/12`
`2)`
\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)
`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)
`\Rightarrow`\(x=\dfrac{40}{21}\)
Vậy, `x = 40/21`
`3)`
\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)
`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)
`\Rightarrow`\(x=\dfrac{16}{21}\)
Vậy, `x = 16/21`
`4)`
\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{1}{12}\)
`\Rightarrow`\(x=\dfrac{1}{12}\div3\)
`\Rightarrow`\(x=\dfrac{1}{36}\)
Vậy, `x = 1/36`
`5)`
\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)
`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{52}{21}\)
Vậy, `x = 52/21`
`6)`
\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)
`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(5x=\dfrac{1}{6}\)
`\Rightarrow`\(x=\dfrac{1}{6}\div5\)
`\Rightarrow`\(x=\dfrac{1}{30}\)
Vậy, `x = 1/30.`
số số hạng là :
( 2500 - 2 ) : 2 + 1 = 1250 ( số )
Tổng dãy trên là :
( 2500 + 2 ) x 1250 : 2 = 1563750
Đ/S : 1563750
a: \(=\dfrac{6x^2-3x+4x^2+2x}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x\left(4x+5\right)}\)
\(=\dfrac{10x^2+x}{\left(2x+1\right)}\cdot\dfrac{2x-1}{2x\left(4x+5\right)}\)
\(=\dfrac{\left(10x^2+x\right)\left(2x-1\right)}{2x\cdot\left(2x+1\right)\left(4x+5\right)}\)
b: \(=\left(\dfrac{x}{\left(5x-1\right)\left(5x+1\right)}\cdot\dfrac{x\left(5x+1\right)}{5x}\right)\cdot\dfrac{x\left(5x+1\right)}{5x-1}+\dfrac{x}{5x-1}\)
\(=\dfrac{x}{5\left(5x-1\right)}\cdot\dfrac{x\left(5x+1\right)}{5x-1}+\dfrac{x}{5x-1}\)
\(=\dfrac{x^2\left(5x+1\right)+5x\left(5x-1\right)}{5\left(5x-1\right)^2}\)
\(=\dfrac{5x^3+x^2+25x^2-5x}{5\left(5x-1\right)^2}=\dfrac{5x^3+26x^2-5x}{5\left(5x-1\right)^2}\)
c: \(=\dfrac{x+1}{x-2}+\dfrac{1-3x}{x\left(x^2+1\right)}\cdot\dfrac{x^2+1}{x-1}\)
\(=\dfrac{x+1}{x-2}+\dfrac{1-3x}{x\left(x-1\right)}\)
\(=\dfrac{x^3-x+\left(1-3x\right)\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\)
\(=\dfrac{x^3-x+x-2-3x^2+6x}{x\left(x-1\right)\left(x-2\right)}=\dfrac{x^3-3x^2+6x-2}{x\left(x-1\right)\left(x-2\right)}\)
\(a.\)
\(F\left(x\right)=2\left(x^4+x^3\right)+2x-4\left(x^2-x^3-1\right)+4\)
\(=2x^4+2x^3+2x-4x^2+4x^3-1\)
\(=4x^4+6x^3-4x^2+2x-1\)
\(G\left(x\right)=5x^4-4\left(3+x^4\right)-2x^2+4x^3+2\left(x^3-x^2+x\right)\)
\(=5x^4-12-4x^4-2x^2+4x^3+2x^3-2x^2+2x\)
\(=x^4+6x^3-4x^2+2x-12\)
\(b.\)
\(K\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(=\left(4x^4+6x^3-4x^2+2x-1\right)+\left(x^4+6x^3-4x^2+2x-12\right)\)
\(=5x^4+12x^3-8x^2+4x-13\)
\(H\left(x\right)=F\left(x\right)-G\left(x\right)\)
\(=\left(4x^4+6x^3-4x^2+2x-1\right)-\left(x^4+6x^3-4x^2+2x-12\right)\)
\(=4x^4+6x^3-4x^2+2x-1-x^4-6x^3+4x^2-2x+12\)
\(=3x^4+11\)
\(c.\)
Ta có : \(H\left(x\right)=36\)
\(\Rightarrow3x^4+11=36\)
\(\Rightarrow3x^4=25\)
a ) x2 - 10x
= x(x - 10)
b) x2 + 2x - 3
= x2 - x + 3x - 3
= x(x - 1) + 3(x - 1)
= (x + 3)(x - 1)
c) x2 + 4 + 4x) = (x + 2)2
LÀM CHO Ý KHÓ NHẤT NHA
\(x^4+1=x^4-x^3+x^2+x^3-x^2+x+x^2-x+1\)
\(=x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x+1\right)^2\)
/x+1/+/x+2/+/x+3/+/x+4/=5x-1
x+1+x+2+x+3+x+4+x=5x-1
[x+x+x+x+x]+[1+2+3+4]=5x-1
5x+10=5x-1
10+1=5x-5x
11=0x
=> ko có giá trị cần tìm
vậy......
có phải bn viết (x+1)+(x+2)+(x+3)+(x++4)+(x+5)=5x-1 phải ko