rút gọn
P=\(\left(5x+2\right).\left(2-5x\right)-\left(x-3\right).\left(x^2+3x+9\right)+x\left(x^2+21x\right)-23\)
rút gọn biểu thức
a) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
b) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
a: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b: \(=3x^2-6x-5x+5x^2-8x^2+24\)
=-11x+24
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.\(\left(-2x^2+5\right)\left(-x+3\right)-x^2\left(2x-6\right)\) \(=\left(x-1\right)\left(x+1\right)-\left(x-2\right)\left(x+4\right)\)
rút gọn biểu thức sau bằng cách nhanh nhất
A = \(\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
B = \(\left(3x^3+3x+1\right)\cdot\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
C = \(\left(2-6x\right)^2+\left(2-5x\right)^2+2\cdot\left(6x-2\right)\cdot\left(2-5x\right)\)
D = \(5\cdot\left(3x-1\right)^2+4\cdot\left(5x+1\right)^2-12\cdot\left(5x-2\right)\left(5x+2\right)\)
E = \(\left(3x-1\right)^2+\left(2x+4\right)\cdot\left(1-3x\right)+\left(x+2\right)^2\)
G = \(\left(x-1\right)^3+4\cdot\left(x+1\right)\cdot\left(1-x\right)+3\cdot\left(x-1\right)\cdot\left(x^2+x+1\right)\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
Tìm x, biết:
\(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)
\(5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(7x\left(x-2\right)-5x\left(x-1\right)=21x^2-14x^2+3\)
Giải gấp giùm mik nha. Cảm ơn nhiều!!!
5x (1/5x -2) + 3(6-1/3x^2) =12
x^2 - 10x + 18 -x^2 =12
-10x + 18 = 12
-10x = -6
x= 6/10
5(x^2 - 3x +1) + x(1-5x) = x-2
5x^2 - 15x + 5 + x - 5x^2 = x-2
-15x = -7
x= 7/15
Rút gọn biểu thức
a, \(A=\left(x+2\right)^2+4\left(x+2\right)\left(x-2\right)+\left(x-4\right)^2\)
b, \(B=\left(3x^2-2x+1\right)\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)
c, \(C=\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)
b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
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:
a/ \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
b/ \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
c/ \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
a) Ta có: \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left(6x-2\right)^2-2\cdot\left(6x-2\right)\left(5x-2\right)+\left(5x-2\right)^2=0\)
\(\Leftrightarrow\left(6x-2-5x+2\right)^2=0\)
\(\Leftrightarrow x^2=0\)
hay x=0
Vậy: x=0
b) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)-5=0\)
\(\Leftrightarrow x^3-6-x^2+4x=0\)
\(\Leftrightarrow4x-6=0\)
\(\Leftrightarrow4x=6\)
hay \(x=\frac{3}{2}\)
Vậy: \(x=\frac{3}{2}\)
c) Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12-2=0\)
\(\Leftrightarrow x^3+3x-15-x^3-27=0\)
\(\Leftrightarrow3x-42=0\)
\(\Leftrightarrow3x=42\)
hay x=14
Vậy: x=14
Tìm \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
rút gọn
a)\(\left(7x-8\right).\left(7x+8\right)-10.\left(2x+3\right)^2+5x.\left(3x-2\right)^24x.\left(x-5\right)^2\)
b) \(\left(3x+7\right)^3-\left(5x-y\right).\left(25x^2+5xy+y^2\right)+\left(x+2y\right)^3\)
a: \(=49x^2-64-10\left(4x^2+12x+9\right)+5x\left(9x^2-12x+4\right)+4x\left(x^2-10x+25\right)\)
\(=49x^2-64-40x^2-120x-90+45x^3-60x^2+20x+4x^3-40x^2+100x\)
\(=49x^3-91x^2-154\)
b: \(=27x^3+189x^2+441x+343-125x^3+y^3+x^3+6x^2y+12xy^2+8y^3\)
\(=-97x^3+189x^2+441x+6x^2y+12xy^2+9y^3+343\)