Tim x.
a. 2007.x{x-2006/7}=0
b. 5{x-2}+3x{2-x}=0
{ viet cach lam gium minh }
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Tìm x thuộc Q
a, 2007.x(x.2006/7)=0
b, 5.(x-2)+3x(2-x)=0
a)\(2007x\left(x-\frac{2006}{7}\right)=0\)
\(\Rightarrow x\left(x-\frac{2006}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-\frac{2006}{7}=0\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}0\\\frac{2006}{7}\end{cases}}\)
b) \(5.\left(x-2\right)+3x\left(2-x\right)=0\)
\(=5\left(x-2\right)-3x\left(x-2\right)=0\)
\(=\left(x-2\right)\left(5-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\5-3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\3x=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{3}\end{cases}}\)
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Tim x biet:
a,10+x / x+17=3 / 4
b'40+x / 77-x=6/7
Nho viet ca cach lam nua nhe ,ai lam day du minh se tick!
Tim cac so nguyen x thoa man:
a, |17x-5|-17x+5|=0
b,|x+2|+|x-5|=7
Cach lam nua nhe
tim x thuoc Z biet:
|x-5|+x+3=16
|x-4|>x+2
||x+5|-4|=3
Cach lam nua nhe
minh chua biet cach lam,cac ban phaiviet cach lam vao minh moi tick
ko bat buoc lam tat ca cac cau
Tim GTNN cua cac bieu thuc:
a,A=|x-1|+|x-2|+|x-3|+|x-4|+...+|x-10|
b,P=|3x-6|+|y2+1|+2015
Cach lam nua nhe
Lam nhanh minh se tick
Tim GTNN cua cac bieu thuc:
a,A=|x-1|+|x-2|+|x-3|+|x-4|+...+|x-10|
b,P=|3x-6|+|y2+1|+2015
Cach lam nua nhe
Lam nhanh minh se tick
Tim x thuoc Z biet:
a, |x-5|+x+3=16
b, |x-5|-7≥9
c, |x-4|>x+2
d, ||x+5|-4|=3
Lam nhanh ho minh
AI LAM NHANH VA CACH LAM DAY DU MINH SE TICK CHO
kHONG CAN LAM HET TAT CA CAC CAU
Tim x thuoc Z biet:
a, |x-5|+x+3=16
b, |x-5|-7≥9
c, |x-4|>x+2
d, ||x+5|-4|=3
khan cap ngay mai minh phai nop roi
lam nhanh va cach lam day du minh se tick
a)x=9
d)x=-12;-6;-4;2
khẩn cấp thì ghi đáp án cho nhanh
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tìm giá trị các đa thức sau
\(A=x^{15}+3x^{14}+5\) biết x+3=0
\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\) biết x= -3
a) \(A=x^{15}+3x^{14}+5\)
\(=x^{14}\left(x+3\right)+5\)
\(=x^{14}.0+5\)
= 5
b) x = -3 => x + 3 = 0
\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)
\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)
\(=\left(x^{2006}.0+1\right)^{2007}\)
\(=1^{2007}=1\)
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\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)
\(A=x^{14}.\left(x+3\right)+5\)
\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)
\(A=0+5\)
\(A=5\) \(\text{Vậy }A=5\text{ với x+3=0}\)
\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)
\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)
\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)
\(B=\left(x^{2006}.0+1\right)^{2007}\)
\(B=\left(0+1\right)^{2007}\)
\(B=1^{2007}\)
\(B=1\) \(\text{Vậy }B=1\text{ với x=-3}\)
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