Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)a)34-26-54=81-64-625=-608
ko hiểu b
2)a)(3x-1)2=(5/6)2=(-5/6)2
+)3x-1=5/6 =>x=11/18
+)3x-1=-5/6 =>x=1/18
b)(x+7)x-11(1-(x-7)23)=0
=>+)(x+7)x-11=0 =>x+7=0 =>x=-7
+)1-(x+7)23=0 =>(x+7)23=1 =>x+7=1 =>x=-6
B1:
a)x=-3/5*9/25 =>x=-27/125
b)x=(4/7)6:(4/7)4 =>x=(4/7)2=16/49
c)(x/4)2=4:(x/2)
(x/4)2=8/x
x2/16=8/x2
x3=128
x=5,039
B2
M=23.10+22.10/23.4+22.11
=230+220/212+222
=230+28+222
=28(222+1+214)
=2
1. Ta có \(|3x-1|=\frac{1}{2}\)
\(\Rightarrow\)\(\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=(\frac{1}{2}+1):3\\x=(-\frac{1}{2}+1):3\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)
Sau đó tự thay x vào đa thức theo 2 trường hợp trên nha
Sai thì thôi nha bn mik cx chưa lm dạng này bh
Câu 1:
\(A\left(x\right)=6x^4-4x^2-3+9x+5x^2-7x-2x^4+4-2x-4x^4\)
\(=\left(6x^4-2x^4-4x^4\right)+\left(-4x^2+5x^2\right)+\left(-7x-2x\right)+9x+\left(-3+4\right)\)
\(=x^2+9x+1\)
Ta có: \(\left|3x-1\right|=\frac{1}{2}\)
TH1: \(3x-1=\frac{1}{2}\Rightarrow3x=\frac{1}{2}+1=\frac{3}{2}\Rightarrow x=\frac{3}{2}:3=\frac{1}{2}\)
\(A\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+9\cdot\frac{1}{2}+1=\frac{1}{4}+\frac{9}{2}+1=\frac{23}{4}\)
TH2: \(3x-1=\frac{-1}{2}\Rightarrow3x=\frac{-1}{2}+1=\frac{1}{2}\Rightarrow x=\frac{1}{2}:3=\frac{1}{6}\)
\(A\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^2+9\cdot\frac{1}{6}+1=\frac{91}{36}\)
Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết 
a) \(\frac{2x-3}{4-x}=\frac{4-x}{2x-3}\)
\(\left(2x-3\right)\left(2x-3\right)=\left(4-x\right)\left(4-x\right)\)
\(\left(2x-3\right)^2=\left(4-x\right)^2\)
\(4x^2-12x+9=16-8x+x^2\)
\(4x^2-12x+9-16+8x-x^2=0\)
\(3x^2-4x-7=0\)
\(3x^2+3x-7x-7=0\)
\(3x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(3x-7\right)=0\)
\(\hept{\begin{cases}x+1=0\\3x-7=0\end{cases}}\)
\(\hept{\begin{cases}x=-1\\x=\frac{7}{3}\end{cases}}\)
a) Ta có \(\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(4^2-2^4\right)\)
=\(\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(16-16\right)\)
=\(\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot0\)=0
b) \(\left(7^{1997}-7^{1995}\right):\left(7^{1994}\cdot7\right)\)
=\(\left(7^{1995}\left(7^2-1\right)\right):7^{1995}\)
=\(7^2-1\)=\(49-1\)=\(48\)
c Giống câu a
a,<=> 145-2x-1=70
<=> 2x=74
<=> x=37
b, <=> 10-4x+15=17
<=> 4x=8
<=> x=2
\(7^2\cdot5-\left(2x+1\right)=630\div9\)
\(49\cdot5-\left(2x+1\right)=70\)
\(245-\left(2x+1\right)=70\)
\(2x+1=245-70\)
\(2x+1=175\)
\(2x=175-1\)
\(2x=174\)
\(x=174\div2\)
\(x=87\)
\(\left(10-4x\right)+120\div2^3=17\)
\(\left(10-4x\right)+120\div8=17\)
\(\left(10-4x\right)+15=17\)
\(10-4x=17-15\)
\(10-4x=2\)
\(4x=10-2\)
\(4x=8\)
\(x=8\div4\)
\(x=2\)