Tìm x bt
a.x- 4/9=-1/2
b. x-5/6=-7/3
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\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)
\(x=\dfrac{7}{35}+\dfrac{-15}{35}\)
\(x=-\dfrac{8}{35}\)
\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{2}\)
\(x=\dfrac{4}{7}:\dfrac{3}{2}\)
\(x=\dfrac{4}{7}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{21}\)
\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)
\(x+\dfrac{3}{4}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}-\dfrac{3}{4}\)
\(x=-\dfrac{23}{12}\)
\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{13}{18}\)
\(x=\dfrac{-5}{9}-\dfrac{13}{18}\)
\(x=\dfrac{-10}{18}-\dfrac{13}{18}\)
\(x=-\dfrac{23}{18}\)
Lời giải:
a.
$x=\frac{7}{25}+\frac{-1}{5}=\frac{7}{25}+\frac{-5}{25}=\frac{7-5}{25}=\frac{2}{25}$
b.
$x=\frac{5}{11}+\frac{4}{-9}=\frac{5}{11}-\frac{4}{9}=\frac{45}{99}-\frac{44}{99}=\frac{1}{99}$
c.
$\frac{x}{-1}=\frac{-1}{3}-\frac{5}{9}=\frac{-3}{9}-\frac{5}{9}=\frac{-8}{9}$
$x=(-1).\frac{-8}{9}=\frac{8}{9}$
a)\(\dfrac{x}{60}=-\dfrac{3}{4}\)
\(\Rightarrow x\cdot4=60\cdot\left(-3\right)\)
\(x\cdot4=-180\)
x=45
b)\(\dfrac{2}{5}=\dfrac{12}{x}\)
\(\Rightarrow2x=5\cdot12\)
\(2x=60\)
x=30
c)\(x-\dfrac{5}{7}=\dfrac{6}{21}\)
\(x=\dfrac{2}{7}+\dfrac{5}{7}\)
x=1
d)\(x+\dfrac{7}{8}=\dfrac{63}{24}\)
\(x=\dfrac{21}{8}-\dfrac{7}{8}\)
\(\dfrac{14}{8}\)
BÀI 1:
\(a,x^2-2x-1\)
\(=x^2-2x+1-2\)
\(=\left(x-1\right)^2-2\)
Vì: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2-2\ge-2\forall x\)
Dấu = xảy ra khi : \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy: GTNN của bt là -2 tại x=1
\(b,4x^2+4x-5\)
\(=4x^2+4x+1-6\)
\(=\left(2x+1\right)^2-6\)
Vì: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2-6\ge-6\forall x\)
Dấu = xảy ra khi \(\left(2x+1\right)^2=0\Rightarrow x=-\frac{1}{2}\)
VậyGTNN của bt là -6 tại x=-1/2
BÀI 2:
\(a,2x-x^2-4\)
\(=-x^2+2x-4\)
\(=-x^2+2x-1-3\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Vì: \(-\left(x-1\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-1\right)^2-3\le-3\forall x\)
Dấu = xảy ra khi : \(-\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy GTLN của bt là -3 tại x=1
b,mk chưa nghĩ ra,lúc nào mk nghĩ ra sẽ gửi lời giải cho bn
1)
a) Đặt \(A=x^2-2x+1\)
\(\Rightarrow A=x^2-2x-1=\left(x^2-2.x.1+1^2\right)-2=\left(x-1\right)^2-2\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\Rightarrow\left(x-1\right)^2-2\ge2\forall x\)
\(A=2\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy \(A_{min}=2\Leftrightarrow x=1\)
Câu b tương tự
2)
a) Đặt \(B=2x-x^2-4\)
\(B=2x-x^2-4=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\Rightarrow-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-3\le-3\forall x\)
\(B=-3\Leftrightarrow-\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy\(B_{max}=-3\Leftrightarrow x=1\)
b) Đặt \(C=-x^2-4\)
Ta có: \(x^2\ge0\forall x\Rightarrow-x^2\ge0\forall x\Rightarrow-x^2-4\le-4\forall x\)
\(C=-4\Leftrightarrow-x^2=0\Leftrightarrow x=0\)
Vậy \(C_{max}=-4\Leftrightarrow x=0\)
a. \(x-\frac{3}{7}:\frac{9}{14}=-\frac{7}{6}\Rightarrow x-\frac{2}{3}=-\frac{7}{6}\Rightarrow x=-\frac{1}{2}\)
b. \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{8}\Rightarrow\frac{1}{4}x=-\frac{1}{8}\Rightarrow x=-\frac{1}{2}\)
c. \(\left|4x-1\right|-\frac{1}{2}=3\Rightarrow\left|4x-1\right|=\frac{7}{2}\Leftrightarrow\orbr{\begin{cases}4x-1=\frac{7}{2}\\4x-1=-\frac{7}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{8}\\x=-\frac{5}{8}\end{cases}}}\)
d. \(25\%x+x=-1,25\Rightarrow125\%x=-1,25\Rightarrow\frac{5}{4}x=-\frac{5}{4}\Rightarrow x=-1\)
B1:
A.1+1=2
B.6+2+5=13
C.6+4+6+2+7=25
D.6+4+7+8=25
B2:
A.x+2=3
x=3-2
x=1
B.x+5=4+6
x+5=10
x=10-5
x=5
C.7=4+x
x=7-4
x=3
D.4+x={3+5+7}+{3+5}
4+x=15+8
4+x=23
x=23-4
x=19
Bài 1 :
A.1+1=2
B.6+2+5=13
C.6+4+6+2+7=25
D.6+4+7+8=25
Bài 2 :
A.x+2=3
x=3-2
x=1
B.x+5=4+6
x+5=10
x=10-5
x=5
C.7=4+x
x=7-4
x=3
D.4+x={3+5+7}+{3+5}
4+x=15+8
4+x=23
x=23-4
x=19
Chúc bạn học tốt nha ~_~
a) \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Rightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy x = 2
b) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Rightarrow-3x=-48\)
\(\Leftrightarrow x=16\)
Vậy x = 16
c) \(\dfrac{1}{9}=\dfrac{-2x}{10}\)
\(\Rightarrow-18x=10\)
\(\Leftrightarrow x=-\dfrac{5}{9}\)
Vậy \(x=-\dfrac{5}{9}\)
d) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{x}-5=\dfrac{-9}{x}+2\)
\(\Leftrightarrow\dfrac{3-5x}{x}=\dfrac{-9+2x}{x}\)
\(\Rightarrow3-5x=-9+2x\)
\(\Leftrightarrow7x=12\)
\(\Leftrightarrow x=\dfrac{12}{7}\)
Vậy \(x=\dfrac{12}{7}\)
e) ĐKXĐ: \(x\ne0\)
\(\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Rightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy \(x=\pm4\)
a) Ta có: \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Leftrightarrow-x=\dfrac{6\cdot\left(-8\right)}{3}=-16\)
hay x=16
Vậy: x=16
a) \(x-\dfrac{4}{9}=\dfrac{-1}{2}=>x=\dfrac{-1}{2}+\dfrac{4}{9}=>x=\dfrac{-9}{18}+\dfrac{8}{18}=>x=-\dfrac{1}{18}\)
b) \(x-\dfrac{5}{6}=-\dfrac{7}{3}=>x=\dfrac{-7}{3}+\dfrac{5}{6}=>x=-\dfrac{14}{6}+\dfrac{5}{6}=>x=-\dfrac{3}{2}\)
Chúc bn học tốt nha
\(a,x-\dfrac{4}{9}=\dfrac{-1}{2}\)
\(\Rightarrow x=\dfrac{-1}{2}+\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{-1}{18}\)
\(b,x-\dfrac{5}{6}=\dfrac{-7}{3}\)
\(\Rightarrow x=\dfrac{-7}{3}+\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{-3}{2}\)