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10. Tatua x katika mlinganyo: Tunapewa mlinganyo: (1)/(2) _2 81 + _2 (x^2 - (x)/(3)) = 1 Step 1: Rahisisha neno la kwanza kwa kutumia sheria ya logariti a^b = b a. (1)/(2) _2 81 = _2 (81^1/2) = _2 sqrt(81) = _2 9 Mlinganyo unakuwa: _2 9 + _2 (x^2 - (x)/(3)) = 1 Step 2: Unganisha maneno ya logariti kwa kutumia sheria a + b = (ab). _2 (9 (x^2 - (x)/(3))) = 1 _2 (9x^2 - 3x) = 1 Step 3: Badilisha kutoka fomu ya logariti kwenda fomu ya kielelezo. 9x^2 - 3x = 2^1 9x^2 - 3x = 2 Step 4: Panga upya mlinganyo kuwa mlinganyo wa quadratic. 9x^2 - 3x - 2 = 0 Tumia fomula ya quadratic x = -b ± sqrt(b^2 - 4ac)2a, ambapo a=9, b=-3, c=-2. x = -(-3) ± sqrt((-3)^2 - 4(9)(-2))2(9) x = 3 ± sqrt(9 + 72)18 x = 3 ± sqrt(81)18 x = (3 ± 9)/(18) Step 5: Tafuta thamani za x. x_1 = (3 + 9)/(18) = (12)/(18) = (2)/(3) x_2 = (3 - 9)/(18) = (-6)/(18) = -(1)/(3) Step 6: Hakikisha suluhisho ni halali kwa kikoa cha logariti (x^2 - (x)/(3) > 0). Kwa x = (2)/(3): ((2)/(3))^2 - (1)/(3)((2)/(3)) = (4)/(9) - (2)/(9) = (2)/(9) > 0. Hii ni halali. Kwa x = -(1)/(3): (-(1)/(3))^2 - (1)/(3)(-(1)/(3)) = (1)/(9) + (1)/(9) = (2)/(9) > 0. Hii pia ni halali. Thamani za x ni: x = (2)/(3), -(1)/(3) 11. Andika upanuzi wa (2 - (1)/(5)x)^4 hadi neno la x^4. Tutatumia Theorem ya Binomial (a+b)^n = _k=0^n nk a^n-k b^k. Hapa a=2, b=-(1)/(5)x, na n=4. Step 1: Tafuta kila neno la upanuzi. Neno la kwanza (k=0): 40 (2)^4-0 (-(1)/(5)x)^0 = 1 · 2^4 · 1 = 16 Neno la pili (k=1): 41 (2)^4-1 (-(1)/(5)x)^1 = 4 · 2^3 · (-(1)/(5)x) = 4 · 8 · (-(1)/(5)x) = -(32)/(5)x Neno la tatu (k=2): 42 (2)^4-2 (-(1)/(5)x)^2 = 6 · 2^2 · ((1)/(25)x^2) = 6 · 4 · (1)/(25)x^2 = (24)/(25)x^2 Neno la nne (k=3): 43 (2)^4-3 (-(1)/(5)x)^3 = 4 · 2^1 · (-(1)/(125)x^3) = 8 · (-(1)/(125)x^3) = -(8)/(125)x^3 Neno la tano (k=4): 44 (2)^4-4 (-(1)/(5)x)^4 = 1 · 2^0 · ((1)/(625)x^4) = 1 · 1 · (1)/(625)x^4 = (1)/(625)x^4 Step 2: Andika upanuzi kamili. (2 - (1)/(5)x)^4 = 16 - (32)/(5)x + (24)/(25)x^2 - (8)/(125)x^3 + (1)/(625)x^4 Sehemu ya pili ya swali haionekani wazi kwenye picha. Tuma swali linalofuata 📸