Tìm x :
a, \(\left(2x+1\right)^2-3x^2+4=\left(1-x\right).\left(1+x\right)\)
b, \(\left(4x-3\right)\cdot\left(4x+3\right)-2\cdot\left(x+2\right)^2=14x^2\)
c, \(\left(2x-1\right)\cdot\left(x+1\right)-x^2+1=\dfrac{1}{2}\cdot\left(x-1\right)^2\)
Tìm x :
\(3x\cdot\left(x-2\right)-2x\cdot\left(2x-1\right)=\left(1-x\right)\cdot\left(1+x\right)\)
\(\left(5x+3\right)\cdot\left(3x-5\right)-\left(x-2\right)\cdot\left(2x+1\right)=6x\cdot\left(3x+1\right)-x^2\)
\(\left(2x-1\right)\cdot\left(2x+1\right)-3\cdot\left(x-1\right)=\left(1-4x\right)\cdot\left(1-x\right)\)
\(\left(2x^2+1\right)\cdot\left(3x^2-1\right)-\left(4x^2-3\right)\cdot\left(x^2+1\right)=x\cdot\left(2x^3+1\right)\)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
Tìm x :
a, \(4x^2-\left(3x+1\right)\cdot\left(2x-1\right)=2\cdot\left(x-3\right)^2\)
b.\(\left(5x-1\right)\cdot\left(x+1\right)-\left(2x-1\right)\cdot\left(2x+1\right)=x\cdot\left(x+1\right)\)
c, \(7x^2-\left(2x-3\right)^2=1+3\cdot\left(x+2\right)^2\)
\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)
\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)
\(\Leftrightarrow-4x^2+13x-17=0\)
\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)
\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)
\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm
\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)
\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)
\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)
\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)
Vậy phương trình vô nghiệm
Chứng minh biểu thức không phụ thuộc x :
1, \(\left(3x-1\right)^2-2\cdot\left(2x-3\right)\cdot\left(2x+3\right)-\left(x-3\right)^2\)
2, \(\left(3x+2\right)^3-\left(3x-2\right)^3-3\cdot\left(6x-1\right)\cdot\left(6x+1\right)\)
3, \(\left(3x-5\right)^2+3\cdot\left(x+1\right)\cdot\left(x-1\right)-\left(4x-3\right)^2+\left(2x+2\right)\cdot\left(2x+1\right)\)
Làm giúp mình nha
\(a,\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x^2-10\right)=72\)
\(b,\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)\cdot\left(x+4\right)\cdot\left(x+5\right)=40\)
\(c,\left(2x-5\right)^2=\left(4x+7\right)^2\)
\(d,\left(2x^2+3x-1\right)^2-5\cdot\left(2x^2+3x+3\right)+24=0\)
a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)
b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x2+5x+4)(x2+5x+6)(x+5)=40
câu c dễ lắm... bạn có thể làm 2 cách 1 là xét 2 th khi vế trái bằng vế phải hay trái dấu vế phải
hoặc cách 2 đưa về hiệu 2 bình phương nhá.. cách này dễ hơn
Chứng minh giá trị biểu thức không phụ thuộc x :
1, \(\left(2x+1\right)^3-\left(2x-1\right)^3-2\cdot\left(4x+3\right)^2+8\cdot\left(x+3\right)^2\)
2,\(\left(2x+1\right)^2\cdot\left(x-1\right)-2\cdot\left(x-2\right)^3+x\cdot\left(3-2x\right)\cdot\left(3+x\right)-\left(3x-3\right)^2\)
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
Tìm x :
\(\left(2x^2+1\right)\cdot\left(3x^2-1\right)-\left(4x^2-3\right)\cdot\left(x^2+1\right)=x\cdot\left(2x^2+1\right)\)
Sửa đề: \(\left(2x^2+1\right)\left(3x^2-1\right)-\left(4x^2-3\right)\left(x^2+1\right)=x^2\left(2x^2+1\right)\)
\(\Leftrightarrow6x^4-2x^2+3x^2-1-\left(4x^4+4x^2-3x^2-3\right)=2x^4+x^2\)
\(\Leftrightarrow6x^4+x^2-1-4x^4-x^2+3-2x^4-x^2=0\)
\(\Leftrightarrow-x^2+2=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
\(P=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right)\cdot\left(\frac{\left(x^3-2x^2-2x-1\right)\cdot\left(x+1\right)}{x^9+x^7-3x^2-3}\right)+1-\frac{2\left(x+6\right)}{x^2+1}\right]\cdot\frac{4x^2+4x+1}{\left(x+3\right)\left(4-x\right)}\)
a, Tìm ĐKXD của P
b,Rút Gọn P
c,Chứng Minh Với các giá trị của x mà biểu thức P có nghĩa thì \(-5\le P\le0\)
Tính:
\(a)\left(-2x^2\right)\cdot\left(3x-4x^3+7-x^2\right)\)
\(b)\left(x+3\right)\cdot\left(2x^2-3x-5\right)\)
\(c)\left(-6x^5+7x^4-6x^3\right):3x^3\)
\(d)\left(9x^2-4\right):\left(3x+2\right)\)
\(e)\left(2x^4-13x^3+15x^2+11x-3\right):\left(x^2-4x-3\right)\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
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