tìm x biết
\(\frac{\left(24-x\right)^2+\left(24-x\right)\left(x-25\right)+\left(x-25\right)^2}{\left(24-x\right)^2-\left(24-x\right)\left(x-25\right)+\left(x-25\right)^2}=\frac{19}{49}\)
so sánh: \(A=26^2-24^2\) và \(B=27^2-25^2\)
tìm x, biết:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
Bài 1:
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)
=>A<B
Bài 2:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
=>4x+13=11
=>4x=-2
=>\(x=-\dfrac{1}{2}\)
tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
Bài 1:
a)\(\left|x-2\right|\)+\(\left|1-\dfrac{x}{2}\right|\)=0 b)\(\left(x-\dfrac{1}{3}\right)^3\)=\(\dfrac{-8}{27}\)
c)\(\dfrac{x^2}{6}\)=\(\dfrac{24}{25}\) c)\(\dfrac{x-1}{x+5}\)=\(\dfrac{6}{7}\)
Giúp mik làm bài này với ạ mik đang cần gấp tối nay mik phải nộp rồi mong mn giúp đỡ mik. Mik cảm ơn mn
Giải phương trình sau:
\(6\left(x-\frac{1}{x}\right)^2-25\left(x-\frac{1}{x}\right)+24=0\)
Giải PT:
a) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
b) \(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
c) \(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
d) \(\sqrt{\dfrac{1}{3x+2}}-\dfrac{1}{2}\sqrt{\dfrac{9}{3x+2}}+\sqrt{\dfrac{16}{3x+2}}-5\sqrt{\dfrac{1}{12x+8}}=1\)
e) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\)
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
d. ĐKXĐ: $x>\frac{-2}{3}$
PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \frac{1}{3x+2}=1$
$\Leftrightarrow 3x+2=1$
$\Leftrightarrow x=-\frac{1}{3}$
Rút gọn biểu thức:
\(A=\left|\frac{\left|y-x\right|}{\left|xy\right|}\right|+\left|\frac{y+x}{xy}-\frac{2}{z}\right|+\frac{\left|y-x\right|}{\left|xy\right|}+\frac{y+x}{xy}+\frac{2}{z}\)
với \(x>5\); \(y=\frac{x^2-25}{x+\frac{10x+25}{x}}\); \(z=\frac{x^2-25}{x+\frac{15x+25}{x-5}}\)
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(\frac{17-x}{12}=\frac{3}{17-x}\)
d) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)=0\)0
b) \(\left(\frac{x}{3}-5\frac{1}{4}\right)^2-\frac{-2}{5}=1\frac{1}{25}\)
c) \(1\frac{1}{3}-25\%\left(x-\frac{8}{3}\right)+2x=1,6:\frac{3}{5}\)
Các bạn giúp mk với, nhanh nhé, mk cần gấp
Câu a có 1 số 0 to thôi nhé!
Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(b,\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)