TÌM X BIẾT
\(x.\left(2+x\right)+\left(4x+4\right)+\left(7x+6\right)+...+\left(28x+20\right)=1560\)
AI LÀM ĐƯỢC CHO 3 TIK
PHẢI GIẢI RA ĐẦY ĐỦ NHA !!!
Tim \(x\in Z\), biet rang :
a) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
b) \(\left(x+2\right)+\left(4x+4\right)+\left(7x+6\right)+...+\left(25x+28\right)+\left(28x+20\right)=1560\)
a, Vì trong mỗi ngoặc có một số hạng nên vì có 100 số hạng nên có 100x
ta có 100x+(1+2+3+.....+100)=5750
100x+5050=5750
100x=5750-5050
100x=700
x=700:100
x=7
nếu tính ko nhầm
a)(x+1)+(x+2)+...+(x+100)=5750
(x+x+...+x)+(1+2+...+100)=5750
1+2+...+100 có: (100-1)+1 =100 số hạng
1+2+...+100=(100+1)*100/2=5050
=>100x+5050=5750
100x=5750-5050
100x=700
x=700/100
x=7. Vậy x=7
Ta có:100x+(1+2+3+...+100)=5750
=>100x+5050=5750
=>100x=5750-5050
=>100x=700
=>x=700:100
=>x=7
Vậy x=7
tìm x, biết:
\(a,-4x\left(x-7\right)+4x\left(x^2-5\right)\)\(=28x^2-13\)
\(b,\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)\))\(=\left(-4+x\right)\left(-2x+3\right)+12x^2+2x^2\)
a) -4x(x - 7) + 4x(x2 - 5) = 28x2 - 13
=> -4x2 + 28x + 4x2 - 20x = 28x2 - 13
=> (-4x2 + 4x2) + (28x - 20x) = 28x2 - 13
=> 8x = 28x2 - 13
=> 8x - 28x2 + 13 = 0
=> phương trình vô nghiệm
b) (4x2 - 5x)(3x + 2) - 7x(x + 5) = (-4 + x)(-2x - 3) + 12x2 + 2x2
=> 4x2(3x + 2) - 5x(3x + 2) - 7x2 - 35x = -4(-2x - 3) + x(-2x - 3) + 14x2
=> 12x3 + 8x2 - 15x2 - 10x - 7x2 - 35x = 8x + 12 - 2x2 - 3x + 14x2
=> 12x3 + (8x2 - 15x2 - 7x2) + (-10x - 35x) = (8x - 3x) + 12 + (-2x2 + 14x2)
=> 12x3 - 14x2 - 45x = 5x + 12 + 12x2
=> 12x3 - 14x2 - 45x - 5x - 12 - 12x2 = 0
=> 12x3 + (-14x2 - 12x2) + (-45x - 5x) - 12 = 0
=> 12x3 - 26x2 - 50x - 12 = 0
Làm nốt
Cái câu b sửa cái đề lại nhé dấu " = " ở chỗ (-2x = 3) là gì vậy?
a, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3+13=0\)( vô nghiệm )
tìm x, biết:
\(a,-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(b,\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)\) \(=\left(-4+x\right)\left(-2x+3\right)+12x^2+2x^2\)
Tìm x biết :(đề không sai )
1.\(-4x\left(x-7\right)+4x\left(x^2-5\right)\) \(=28x^2-13\)
2.\(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)\) \(=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
3.\(\left(-4x^2-3\right)\left(2x+5\right)\) \(-\left(8x-3\right)\left(-x^2+2\right)\) \(-5x\left(x-6\right)-3x^2-4\)
4.\(\left(x-7\right)\left(x+5\right)-\left(x-3\right)\left(x-2\right)\) \(=15x^2\left(x+1\right)-\left(3x^2-1\right)\) \(\left(5x^2-2\right)-21x^2\)
5.\(\left(x-3\right)\left(-x+10\right)+\left(x-8\right)\left(x+3\right)\)\(=\left(5x^2-1\right)\left(x+3\right)-5x^3-15x^2\)
6.\(\left(-2x^2+5\right)\left(-x+3\right)-x^2\left(2x-6\right)\) \(=\left(x-1\right)\left(x+1\right)-\left(x-2\right)\left(x+4\right)\)
1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )
2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)
\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)
\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )
Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))
1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
=> \(-4x^2+28x+4x^3-20x=28x^2-13\)
=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)
=> \(-4x^2+4x^3+8x-28x^2+13=0\)
=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)
=> \(-32x^2+4x^3+8x+13=0\)
=> vô nghiệm
2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)
=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)
=> \(-14x^2-56x+12=0\)
=> .... tự tìm
Câu c dấu bằng chỗ nào ?
\(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(< =>-4x^2+28x+4x^3-20x=28x^2-13\)
\(< =>4x^3-\left(4x^2+28x^2\right)+8x+13=0\)
\(< =>4x^3-32x^2+8x+13=0\)
do cái này nghiệm màu mè nên mình sẽ làm cách khá khó hiểu
\(< =>x^3-7x^2+2x+\frac{13}{4}=0\)
Đặt \(x=y+\frac{7}{3}\)khi đó phương trình trở thành
\(\left(y+\frac{7}{3}\right)^3-7\left(y+\frac{7}{3}\right)^2+2\left(y+\frac{7}{3}\right)+\frac{13}{4}=0\)
\(< =>y^3+3y^2\frac{7}{3}+3y\frac{49}{9}-7\left(y^2+\frac{14y}{3}+\frac{49}{9}\right)+2y+\frac{14}{3}+\frac{13}{4}=0\)
\(< =>y^3+7y^2+\frac{49y}{3}-7y^2-\frac{98y}{3}-\frac{343}{9}+2y+\frac{95}{12}=0\)
\(< =>y^3-\frac{43y}{3}-\frac{1087}{36}=0\)
Đặt \(y=u+v\)sao cho \(uv=\frac{43}{9}\)khi đó pt trở thành
\(\left(u+v\right)^3-\frac{43\left(u+v\right)}{3}-\frac{1087}{36}=0\)
\(< =>u^3+v^3+3uv\left(u+v\right)-\left(u+v\right).\frac{43}{3}-\frac{1087}{36}=0\)
\(< =>u^3+v^3+\left(u+v\right)\left(3uv-\frac{43}{3}\right)-\frac{1087}{36}=0\)
\(< =>u^3+v^3=\frac{1087}{36}\)(*) (do \(uv=\frac{43}{9}\Leftrightarrow3uv-\frac{43}{3}=0\))
Ta có \(uv=\frac{43}{9}\Leftrightarrow u^3v^3=\frac{79507}{729}\)(**)
Từ (*) và (*) Suy ra được \(\hept{\begin{cases}u^3+v^3=\frac{1087}{36}\\u^3v^3=\frac{79507}{729}\end{cases}}\)
đến đây dễ rồi nhé ^^
Tìm x biết : (đề không sai)
1.\(-4x\left(x-7\right)+4x\left(x^2-5\right)\) \(=28x^2-13\)
2.\(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x-7\right)\)= \(\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
3.\(\left(-4x^2-3\right)\left(2x+5\right)-\left(8x-3\right)\) \(\left(-x^2+2\right)=-5x^2\left(x-6\right)-3x^2-4\)
4.\(\left(x-7\right)\left(x+5\right)-\left(x-3\right)\left(x-2\right)\) \(=15x^2\left(x+1\right)-\left(3x^2-1\right)\) \(\left(5x^2-2\right)-21x^2\)
5.\(\left(x-3\right)\left(-x+10\right)+\left(x-8\right)\left(x+3\right)\) \(=\left(5x^2-1\right)\left(x+3\right)-5x^3-15x^2\)
6.\(\left(-2x^2+5\right)\left(-x+3\right)-x^2\left(2x-6\right)\) \(=\left(x-1\right)\left(x+1\right)-\left(x-2\right)\left(x+4\right)\)
Tìm x thuộc N
\(a,\left(x-9\right)^4=\left(x-9\right)^7\)
\(b,\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(c,\left(x-8\right)^3=\left(x-8\right)^6\)
cho mik lời giải nhé mik tik cho là lời giải ko phải mỗi đáp án ko đâu nhế
a) \(\left(x-9\right)^4=\left(x-9\right)^7\)
\(\Rightarrow\left[{}\begin{matrix}x-9=1\\x-9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=10\\x=9\end{matrix}\right.\)
b) \(\left(3x-15\right)^{10}=\left(3x-15\right)^{15}\)
\(\Rightarrow\left[{}\begin{matrix}3x-15=0\\3x-15=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{3}\\x=\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left(x-8\right)^3=\left(x-8\right)^6\)
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x-8=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=8\\x=9\end{matrix}\right.\)
Tính \(7x\left(2-3x\right)+x^2\left(2x+1\right)-2x^2\left(x-2\right)+2x\left(8x-7\right)\)
Giải đầy đủ giúp mình nha !!!!
\(7x\left(2-3x\right)+x^2\left(2x+1\right)-2x^2\left(x-2\right)+2x\left(8x-7\right)=14x-21x^2+2x^3+x^2-2x^3+4^2+16x^2-14x=0\)
Tìm x , biết
\(\dfrac{0,1\left(6\right)+0,\left(3\right)}{0,\left(3\right)+1,1\left(6\right)}\)\(\times\) x =\(0,\left(2\right)\) ( giải thích ra nha )
giúp mik vs 4h 30 ) hc rồi
a) \(^{ }\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
b) \(^{ }8\left(x-2\right)-3\left(x^2-4x-5\right)-5x^2\)
c) \(^{^{ }}\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2