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a: \(\Leftrightarrow x^3-x^2+x-x^3-x^2+m=-2x^2+x+5\)
\(\Leftrightarrow m-2x^2+x=-2x^2+x+5\)
hay m=5
b: \(\Leftrightarrow-x^4-x^3-3x^2=-x^4-x^3-x^2+m\)
\(\Leftrightarrow m=-x^4-x^3-3x^2+x^4+x^3+x^2=-2x^2\)
Tách tách tách :v
$(15-2x)(4x+1)-(13-4x)(2x-3)-(x-1)(x+2)+x^2=52$
$=>(60x+15-8x^2-2x)-(26x-39-8x^2+12x)-(x^2+3x+2)+x^2=52$
$=>60x+15-8x^2-2x-26x+39+8x^2-12x-x^2-3x-2+x^2=52$
$=>(8x^2-8x^2+x^2-x^2)+(60x-2x-26x-12x-3x)+(15+39-2)=52$
$=>17x+52=52$
$=>x=0$
A) (15-2x)(4x+1)-(13-4x)(2x-3)-(x-1)(x+2)+x^2=52
..............bn phân rồi gộp lại để ra kq như dòng dưới nha....
=>19x + 56 = 52
=> 19x = -4
=> x = ‐ 4 / 1 9
NHỚ TK MK ĐÓ
a, (x-1).(x-2).(x-3)
= (x2 - 2x - x + 2) . (x-3)
= (x2 - 3x + 2). (x-3)4
= x3 - 3x2 - 3x2 + 9x + 2x -6
= x3 - 6x2 + 11x -6
b) (x2 +x+1)(x2-1)(x2-x+1)
= (x4 - x2 + x3 - x+ x2 -1) . (x2 - x +1)
= (x4 + x3 -x -1) . (x2 - x +1)
= x6 - x5 + x4 + x5 - x4 + x3 - x2 + x -1
= x6 + x3 - x2 + x - 1
c) (2x-5)(4-3x)-(3x+11)(5-2x)-15(2x-5)
= (8x - 6x2 - 20 + 15x) - (15x-6x+55-22x) - 30x + 75
= 8x - 6x2 - 20 + 15x - 15x+6x-55+22x - 30x+75
= 6x-6x2 +55
d)(x2-2x+3)(3x-5)-(x2+x-1)(2x+7)
làm tương tự phần C
lưu ý trước dấu ngoặc là dấu trừ, khi phá ngoặc ra phải đổi dấu
a: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(=\left(x^2-3x+2\right)\left(x-3\right)\)
\(=x^3-3x^2-3x^2+9x+2x-6\)
\(=x^3-6x^2+11x-6\)
b: \(\left(x^2+x+1\right)\left(x^2-1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\cdot\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
c: \(=8x-6x^2-20+15x-\left(15x-6x^2+55-10x\right)-30x+75\)
\(=-6x^2-7x+55+6x^2-5x-55\)
\(=-12x\)
d: \(\left(x^2-2x+3\right)\left(3x-5\right)-\left(x^2+x-1\right)\left(2x+7\right)\)
\(=3x^3-5x^2-6x^2+10x+9x-10-\left(x^2+x-1\right)\left(2x+7\right)\)
\(=3x^3-11x^2+19x-10-\left(2x^3+7x^2+2x^2+7x-2x-7\right)\)
\(=3x^3-11x^2+19x-10-2x^3-9x^2-5x+7\)
\(=x^3-20x^2+14x-3\)
Tìm x
a) Ta có: \(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)-15=0\)
\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)
\(\Leftrightarrow40x-40=0\)
\(\Leftrightarrow40x=40\)
hay x=1
Vậy: x=1
b) Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1\right)-49=0\)
\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)
\(\Leftrightarrow12x-36=0\)
\(\Leftrightarrow12x=36\)
hay x=3
Vậy: x=3
d) Ta có: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(\Leftrightarrow12x-5=0\)
\(\Leftrightarrow12x=5\)
hay \(x=\frac{5}{12}\)
Vậy: \(x=\frac{5}{12}\)
e) Ta có: \(\left(x-5\right)^2-x\left(x-4\right)=9\)
\(\Leftrightarrow x^2-10x+25-x^2+4x-9=0\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow6x=16\)
hay \(x=\frac{8}{3}\)
Vậy: \(x=\frac{8}{3}\)
f) Ta có: \(\left(x-5\right)^2-\left(x-4\right)\left(1-x\right)=0\)
\(\Leftrightarrow x^2-10x+25-\left(x-x^2-4+4x\right)=0\)
\(\Leftrightarrow x^2-10x+25-x+x^2+4-4x=0\)
\(\Leftrightarrow2x^2-15x+29=0\)
\(\Leftrightarrow2\left(x^2-\frac{15}{2}x+\frac{29}{2}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{15}{4}+\frac{225}{16}+\frac{7}{16}=0\)
\(\Leftrightarrow\left(x-\frac{15}{4}\right)^2+\frac{7}{16}=0\)(vô lý)
Vậy: x∈∅
