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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?
1. (3x - 5)2 - (3x + 1)2 = 8
=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8
=> -6(6x - 4) = 8
=> 6x - 4 = \(\dfrac{-4}{3}\)
\(\Rightarrow x=\dfrac{4}{9}\)
2) 2x(8x - 3) - (4x - 3)2 = 27
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> x = 2
3) (2x - 3)2 - (2x + 1)2 = 3
=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3
=> -4(4x - 2) = 3
=> 4x - 2 = \(\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{5}{16}\)
4) (x + 5)2 - x2 = 45
=> (x + 5 - x)(x + 5 + x) = 45
=> 5(2x + 5) = 45
=> 2x + 5 = 9
=> x = 2
5) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 18
=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9(x2 + 2x + 1) = 18
=> -9x2 + 27x + 9x2 + 18x + 9 = 18
=> 45x + 9 = 18
=> 45x = 9
=> x = \(\dfrac{1}{5}\)
6) x(x - 4)(x + 4) - (x - 5)(x2 + 5x + 25) = 13
=> x (x2 - 16) - (x3 - 125) = 13
=> x3 - 16x - x3 + 125 = 13
=> -16x = -112
=> x = 7.
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
\(A=x^2-4x-x\left(x-4\right)-15\)
\(=x^2-4x-x^2+4x-15=-15\) => đpcm
\(B=5x\left(x^2-x\right)-x^2\left(5x-5\right)-13\)
\(=5x^3-5x^2-5x^3+5x^2-13=-13\) => đpcm
\(C=-3x\left(x-5\right)+3\left(x^2-4x\right)-3x+7\)
\(=-3x^2+15x+3x^2-12x-3x+7=7\) => đpcm
\(D=7\left(x^2-5x+3\right)-x\left(7x-35\right)-14\)
\(=7x^2-35x+21-7x^2+35x-14=7\) => đpcm
\(E=4x\left(x^2-7+2\right)-4\left(x^3-7x+2x-5\right)\)
\(=4x^3-20x-4x^3+20x+20=20\) => đpcm
\(H=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x=-10\) => đpcm
\(3x^4-4x^3+2x\left(x^3-2x^2+7x\right)\)
\(=3x^4-4x^3+2x^4-4x^3+14x^2\)
\(=5x^4-8x^3+14x^2\)
3x4 - 4x3 + 2x(x3 - 2x2 + 7x )
= 3x4 - 4x3 + 2x4 _ 4x3 + 14x2
= 5x4 - 8x3 + 14x2
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
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 ^^
4. (x - 7)(x + 5) - (x - 3)(x - 2) = 15x2(x + 1) - (3x2 - 1)(5x2 - 2) - 21x2
=> x(x + 5) - 7(x + 5) - x(x - 2) + 3(x - 2) = 15x3 + 15x2 - 3x2(5x2 - 2) - 1(5x2 - 2) - 21x2
=> x2 + 5x - 7x - 35 - x2 + 2x + 3x - 6 = 15x3 + 15x2 - 15x4 + 6x2 - 5x2 + 2 - 21x2
=> x2 + 5x - 7x -35 - x2 + 2x + 3x - 6 - 15x3 - 15x2 + 15x4 - 6x2 + 5x2 - 2 + 21x2 = 0
=> (x2 - x2 - 15x2 + 5x2 + 21x2 - 6x2) + (5x - 7x + 2x + 3x) + (-35 - 6 - 2) - 15x3 + 15x4 = 0
=> 5x2 + 3x - 43 - 15x3 + 15x4 ( tự tìm x)
5. (x - 3)(- x + 10) + (x - 8)(x + 3) = (5x2 - 1)(x + 3) - 5x3 - 15x2
=> x(-x + 10) - 3(-x + 10) + x(x + 3) - 8(x + 3) = 5x2(x + 3) - 1(x + 3) - 5x3 - 15x2
=> -x2 + 10x + 3x2 - 30 + x2 + 3x - 8x - 24 = 5x3 + 15x2 - x - 3 - 5x3 - 15x2
=> -x2 + 10x + 3x2 - 30 + x2 + 3x - 8x - 24 - 5x3 - 15x2 + x + 3 + 5x3 + 15x2 = 0
=> (-x2 + 3x2 + x2 - 15x2 + 15x2) + (10x + 3x - 8x + x) + (-30 - 24 + 3) + (-5x3 + 5x3) = 0
=> 3x2 + 6x - 51 = 0
=> tự tìm
Làm nốt bài 6 đi
P/S : Bài tìm x nào cũng là căn , vô nghiệm :v
\(\left(-2x^2+5\right)\left(-x+3\right)-x^2\left(2x-6\right)=\left(x+1\right)\left(x-1\right)-\left(x-2\right)\left(x+4\right)\)
\(< =>\left(5-2x^2\right)\left(3-x\right)-2x^3+6x^2=x^2-1-\left(x^2+4x-2x-8\right)\)
\(< =>15-5x-6x^2+2x^3-2x^3+6x^2=x^2-1-x^2-4x+2x+8\)
\(< =>15-5x=-2x+7< =>-5x+2x=7-15\)
\(< =>-3x=-8< =>x=\frac{8}{3}\)
tí nữa mình giải tiếp cho câu 4 luôn nhé ^^
\(15x^4-15x^3+5x^2+3x-43=0\)
\(< =>x^4-x^3+\frac{1}{3}x^2+\frac{1}{5}x-\frac{43}{15}=0\)
\(< =>x^4-x^3=-\frac{1}{3}x^2-\frac{1}{5}x+\frac{43}{15}\)
\(< =>x^4-2.\frac{1}{2}x.x^2+\frac{1}{4}x^2=\left(\frac{1}{4}-\frac{1}{3}\right)x^2-\frac{1}{5}x+\frac{43}{5}\)
\(< =>\left(x^2-\frac{1}{2}x\right)^2=\frac{-1}{12}x^2-\frac{1}{5}x+\frac{43}{5}\)
\(< =>\left(x^2-\frac{x}{2}\right)^2+2\left(x^2-\frac{x}{2}\right).\frac{y}{2}+\frac{y^2}{4}=\left(x^2-\frac{x}{2}\right)y+\frac{y^2}{4}-\frac{1}{12}x^2-\frac{1}{5}x+\frac{43}{5}\)
\(< =>\left(x^2+\frac{y-x}{2}\right)^2=yx^2-\frac{xy}{2}+\frac{y^2}{4}-\frac{1}{12}x^2-\frac{1}{5}x+\frac{43}{5}\)
\(< =>\left(x^2+\frac{y-x}{2}\right)^2=x^2\left(y-\frac{1}{12}\right)-x\left(\frac{y}{2}-\frac{1}{5}\right)+\frac{y^2}{4}+\frac{43}{5}=f\left(x\right)\)
Xét delta của VP ta có : \(\left(\frac{y}{2}-\frac{1}{5}\right)^2-4\left(\frac{y^2}{4}+\frac{43}{5}\right)\left(y-\frac{1}{12}\right)=0\)
mình tính ra y gần bằng 1/12 nhé :)) nên mình để tạm vậy
Với y = 1/12 thì pt trở thành : \(\left(x^2+\frac{\frac{1}{12}-x}{2}\right)^2=-x\left(\frac{1}{24}-\frac{1}{5}\right)+\frac{1}{576}+\frac{43}{5}\)
\(< =>\left(x^2+\frac{\frac{1}{12}-x}{2}\right)^2=\frac{19x}{120}+\frac{24773}{2880}\)
bạn phân tích VP sao cho thành bình phương là xong