Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mik ko chép lại đề bài nha
a) = (123)2- 12- (36. 46)
= (126-1)- (3.4)6
= 126-1-126
= -1
a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
a) Đặt: x = a- b; y = b - c ; z = c- a
Ta có: x + y + z = 0
=> \(A=x^3+y^3+z^3=3xyz+\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=3xyz\)
=> \(A=3xyz=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) Đặt: \(a=x^2-2x\)
Ta có: \(B=a\left(a-1\right)-6=a^2-a-6=\left(a+2\right)\left(a-3\right)=\left(x^2-2x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x^2-2x+2\right)\left(x+1\right)\left(x-3\right)\)
d) \(D=4\left(x^2+2x-8\right)\left(x^2+7x-8\right)+25x^2\)
Đặt: \(x^2-8=t\)
Ta có: \(D=4\left(t+2x\right)\left(t+7x\right)+25x^2\)
\(=4t^2+36xt+81x^2=\left(2t+9x\right)^2\)
\(=\left(2x^2+9x-16\right)^2\)
e sẽ cố gắng !!!
\(3x-15=2x\left(x-5\right)\)
\(3x-15=2x^2-10x\)
\(3x-15-2x^2+10x=0\)
\(13x-15-2x^2=0\)
\(x\left(13-2x\right)-15=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(f,x\left(2x-7\right)-4x+14=0\)
\(2x^2-7x-4x+14=0\)
\(2x^2-11x+14=0\)
\(x\left(2x-11\right)=-14\)
\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)
a) Ta có F = \(\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
=> 8F = \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}\)
=> 8F = \(\left(3^8-1\right)\left(3^8+1\right)-3^{16}=3^{16}-1-3^{16}=-1\)
=> F = -1/8
b) Ta có G = \(\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)-\frac{2^{24}}{7}\)
=> 7G = 7(23 + 1)(26 + 1)(212 + 1) - 224
=> 7G = (23 - 1)(23 + 1)(26 + 1)(212 + 1) - 224
=> 7G = (26 - 1)(26 + 1)(212 + 1) - 224
=> 7G = (212 - 1)(212 + 1) - 224
=> 7G = 224 - 1 - 224
=> 7G = -1
=> G = -1/7
\(F=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\frac{3^{16}}{8}\)
<=> \(\left(3^2-1\right)F=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-\left(3^2-1\right)\frac{3^{16}}{8}\)
<=> \(8F=\left(3^4-1\right)\left(3^4+1\right)\left(3^8-1\right)-3^{16}\)
<=> \(8F=\left(3^8+1\right)\left(3^8-1\right)-3^{16}\)
<=> \(8F=\left(3^{16}-1\right)-3^{16}=-1\)
<=> F = -1/8
Câu G làm tương tự
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
Ta có : 642 + 128.36 + 362
= 642 + 2.64.36 + 362
= (64 + 36)2
= 1002
= 10000
Đúng 0 Báo cáo sai phạm
Ta có : 642 + 128.36 + 362
= 642 + 2.64.36 + 362
= (64 + 36)2
= 1002
= 10000