rút gọn
\(\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+\right)^2\)
Rút gọn
B=\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
C=\(\left(1+tan^2a\right)\left(1-sin^2a\right)+\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(B\sqrt{2}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)\(=\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|-2=\sqrt{5}+1-\sqrt{5}+1-2=0\Rightarrow B=0\)
\(C=\left(1+\frac{\sin^2a}{\cos^2a}\right)\left(1-\sin^2a\right)+\left(1+\frac{\cos^2a}{\sin^2a}\right)\left(1-\cos^2a\right)\)
\(=\left(1+\frac{\sin^2a}{\cos^2a}\right)\left(\cos^2a\right)+\left(1+\frac{\cos^2a}{\sin^2a}\right)\left(\sin^2a\right)\)
\(=\frac{\sin^2a+\cos^2a}{\cos^2a}.\cos^2a+\frac{\cos^2a+\sin^2a}{\sin^2a}.\sin^2a\)
\(=\frac{1}{\cos^2a}.\cos^2a+\frac{1}{\sin^2a}\sin^2a=2\)
B
Bạn dùng theo công thức này
\(\sqrt{m+n\sqrt{p}};\sqrt{m-n\sqrt{p}}\)
Dùng pt bậc 2
\(a=1;b=-m;c=\frac{\left(n\sqrt{p}\right)^2}{4}\)
Nghiệm x1 ; x2
\(\sqrt{\left(\sqrt{x1}+\sqrt{x2}\right)^2};\sqrt{\left(\sqrt{x1}-\sqrt{x2}\right)^2}\)
\(B=\sqrt{\left(\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right)^2}-\sqrt{2}\)
\(=|\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}|-|\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}|-\sqrt{2}\)
\(=\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}-\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right)-\sqrt{2}\)
\(=2\cdot\sqrt{\frac{1}{2}}-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}=0\)
C.
\(=\frac{1}{cos^2a}\cdot cos^2a+\frac{1}{sin^2a}\cdot sin^2a\)
\(=1+1=2\)
Rút gọn biểu thức:
\(\left(1+tan^2a\right)\left(1-sin^2a\right)+\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(\left(1+\frac{\sin^2}{\cos^2}\right)cos^2-\left(1+\frac{cos^2}{sin^2}\right)sin^2.\)
=> \(\frac{cos^2+sin^2}{cos^2}\left(cos^2\right)-\frac{sin^2+cos^2}{sin^2}\left(sin^2\right)\)
=> 1-1 =0
\(=\frac{1}{cos^2a}\cdot cos^2a+\frac{1}{sin^2a}\cdot sin^2a\)
\(=1+1\)
\(=2\)
Bài 1: Rút gọn:
\(A=\left(2a^2+2a+1\right).\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(B=5.\left(2x-1\right)^2+4.\left(x-1\right).\left(x+3\right)-2.\left(5-3x\right)^2\)
GIÚP MIK VS!!! MIK ĐAG CẦN GẤP
Rút gọn các biểu thức sau:
a/ \(\left(3x-1\right)^2-2\left(2-5x\right)-2\left(x^2^{^{ }}+x-1\right)\left(x-\dfrac{1}{2}\right)\)
b/\(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
c/\(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
d/\(\left(3a-1\right)^2+2\left(9a^2-1\right)\left(3a+1\right)\)
e/\(\left(3x-4\right)^2+\left(4-x\right)^2-2\left(3x-4\right)\left(x-4\right)\)
MK CÂNG GẤP Ạ AI NHANH MK SẼ VOTE Ạ
b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
Rút gọn A = \(\left[\frac{\left(a-1\right)^2}{\left(a-1\right)^2+3a}+\frac{2a^2-4a-1}{a^3-1}+\frac{1}{a+1}\right]:\frac{2a}{3}\)
\(=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{1}{a-1}\right]:\dfrac{2a}{3}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{3}{2a}=\dfrac{3}{2a}\)
Rút gọn : \(\left(\frac{1}{2a-b}+\frac{3b}{b^2-4a^2}-\frac{2}{2a+b}\right):\left(1+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(\left(\frac{1}{2a-b}+\frac{3b}{b^2-4a^2}-\frac{2}{2a+b}\right):\left(1+\frac{4a^2+b^2}{4a^2-b^2}\right)\left(ĐK:2a\ne\pm b\right)\)
\(=\left(\frac{1}{2a-b}-\frac{3b}{\left(2b-b\right)\left(2a+b\right)}-\frac{2}{2a+b}\right):\frac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\frac{2a+b-3b-2\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\cdot\frac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)
\(=\frac{2a+b-3b-4a+2b}{8a^2}=\frac{-2a}{8a^2}=-\frac{1}{4a}\)
Rút gọn biểu thức sau : \(R=\left(\frac{a-2}{2a-2}-\frac{3}{2-2a}-\frac{a^2+2a+3}{2a+2}\right).\left(1-\frac{a-3}{a+1}\right)\)
Bài này mình ra kết quả không gọn lắm, nên muốn tham khảo đáp số của mọi người ạ!
\(\left(\dfrac{a+1}{2a-2}+\dfrac{1}{2-2a^2}\right)\dfrac{2a+2}{a+2}\)
a,tìm điều kiện xácđịnh
b,rút gọn PT
tính giá trị của P khi a=2
a) \(ĐKXĐ:a\ne\pm1\)
b) \(P=\left(\dfrac{a+1}{2a-2}+\dfrac{1}{2-2a^2}\right)\cdot\dfrac{2a+2}{a+2}\)
\(=\left(\dfrac{a+1}{2\left(a-1\right)}+\dfrac{1}{2\left(1-a^2\right)}\right)\cdot\dfrac{2\left(a+1\right)}{a+2}\)
\(=\left(\dfrac{a+1}{2\left(a-1\right)}-\dfrac{1}{2\left(a-1\right)\left(a+1\right)}\right)\cdot\dfrac{2\left(a+1\right)}{a+2}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)-1}{2\left(a-1\right)\left(a+1\right)}\cdot\dfrac{2\left(a+1\right)}{a+2}\)
\(=\dfrac{a^2-1-1}{\left(a-1\right)\left(a+2\right)}\)
\(=\dfrac{a^2-2}{a^2+a-2}\)
Khi a = 2 thì :
\(P=\dfrac{2^2-2}{2^2+2-2}=\dfrac{2}{4}=\dfrac{1}{2}\)
p/s: check lại hộ tui nhá =)))
Rút gọn \(D=\left(\frac{a+1}{2a-2}-\frac{1}{2a^2-2}\right).\frac{2a+2}{a+2}\)
Ta có: \(\frac{a+1}{2a-2}-\frac{1}{2a^2-2}=\frac{\left(a+1\right)^2-1}{2\left(a^2-1\right)}=\frac{a^2+2a+1-1}{2\left(a^2-1\right)}=\frac{a\left(a+2\right)}{2\left(a^2-1\right)}\)
Vậy D=\(\frac{a\left(a+2\right)}{2\left(a^2-1\right)}.\frac{2\left(a+1\right)}{a+2}=\frac{a}{a-1}\)